Download Screen

The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Lecture 8: Special Probability Densities
Assist. Prof. Dr. Emel YAVUZ DUMAN
Int. to Prob. Theo. and Stat & Int. to Probability
İstanbul Kültür University
Faculty of Engineering
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Outline
1
The Uniform Distribution
2
The Normal Distribution
3
The Normal Approximation to the Binomial Distribution
4
The Normal Approximation to the Poisson Distribution
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Outline
1
The Uniform Distribution
2
The Normal Distribution
3
The Normal Approximation to the Binomial Distribution
4
The Normal Approximation to the Poisson Distribution
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Definition 1
A random variable has a uniform distribution and it is refereed to
as a continuous uniform random variable if and only if its
probability density is given by
(
1
for α < x < β,
u(x; α, β) = β−α
0
elsewhere.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Definition 1
A random variable has a uniform distribution and it is refereed to
as a continuous uniform random variable if and only if its
probability density is given by
(
1
for α < x < β,
u(x; α, β) = β−α
0
elsewhere.
The parameters α and β of this probability density are real
constants, with α < β,
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Definition 1
A random variable has a uniform distribution and it is refereed to
as a continuous uniform random variable if and only if its
probability density is given by
(
1
for α < x < β,
u(x; α, β) = β−α
0
elsewhere.
The parameters α and β of this probability density are real
constants, with α < β, and may be pictured as in the figure.
u(x; α, β)
1
β−α
α
β
x
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Theorem 2
Them mean and the variance of the uniform distribution are given
by
1
α+β
and σ 2 = (β − α)2
µ=
2
12
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Theorem 2
Them mean and the variance of the uniform distribution are given
by
1
α+β
and σ 2 = (β − α)2
µ=
2
12
Proof. The mean:
µ = E (X ) =
Z
∞
−∞
xu(x; α, β)dx =
Z
β
α
x
dx
β−α
β
2
2
x2
= β − α = (β − α)(β + α)
=
2(β − α) α 2(β − α)
2(β − α)
β+α
=
2
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The variance:
Z
2
E (X ) =
∞
2
x u(x; α, β)dx =
−∞
Z
β
α
x2
dx
β−α
β
3
3
2
2
= β − α = (β − α)(β + βα + α )
=
3(β − α) α 3(β − α)
3(β − α)
=
x3
β 2 + βα + α2
3
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The variance:
Z
2
E (X ) =
∞
2
x u(x; α, β)dx =
−∞
Z
β
α
x2
dx
β−α
β
3
3
2
2
= β − α = (β − α)(β + βα + α )
=
3(β − α) α 3(β − α)
3(β − α)
=
x3
β 2 + βα + α2
3
β 2 + βα + α2 (β + α)2
−
3
22
2
2
2
2
4β + 4βα + 4α − 3β − 6αβ − 3α
=
12
2
2
β − 2βα + α
(β − α)2
=
=
.
12
12
σ 2 = E (X 2 ) − (E (X ))2 =
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 3
Southwest Arizona State University provides bus service to
students while they are on campus. A bus arrives at the North
Main Street and College Drive stop every 30 minutes between 6
am and 11 pm during weekdays. Students arrive at the bus stop at
random times. The time that a student waits is uniformly
distributed from 0 to 30 minutes.
(a) Draw a graph of this distribution.
(b) Show that the area of this uniform distribution is 1.00.
(c) How long will a student typically have to wait for a bus? In
other words what is the mean waiting time? What is the
standard deviation of the waiting times?
(d) What is the probability a student will wait more than 25
minutes?
(e) What is the probability a student will wait between 10 and 20
minutes?
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Solution. (a) The graph of this distribution:
Probability
1
30−0
= 0.03̄
0
10
20
30
Length of wait (minutes)
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
(b) The times students must wait for the bus is uniform over the
interval from 0 minutes to 30 minutes, so in this case α = 0 and
β = 30.
Area = (height)(base)
1
(30 − 0) = 0
(30 − 0)
or
Z
30
1
dx
30 − 0
0
x 30 30 − 0
=
= 1.
=
30 0
30
Area =
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
(c) The mean waiting time:
µ=
α+β
0 + 30
=
= 15
2
2
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
(c) The mean waiting time:
µ=
α+β
0 + 30
=
= 15
2
2
The standard deviation of the waiting time:
r
r
√
(β − α)2
(30 − 0)2
=
= 5 3 = 8.6603
σ=
12
12
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
(c) The mean waiting time:
µ=
α+β
0 + 30
=
= 15
2
2
The standard deviation of the waiting time:
r
r
√
(β − α)2
(30 − 0)2
=
= 5 3 = 8.6603
σ=
12
12
P(X )
0.03̄
0
10
20
µ = 15
30
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
(d) P(25 < X < 30) =
R 30
1
25 30−0 dx
=
x 30
30 25
=
5
30
=
1
6
= 0.16̄
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
(d) P(25 < X < 30) =
R 30
1
25 30−0 dx
P(X )
=
x 30
30 25
=
5
30
=
1
6
Area= 0.16̄
0.03̄
0
10
20 25 30
µ = 15
= 0.16̄
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
(e) P(10 ≤ X ≤ 20) =
R 20
1
10 30−0 dx
=
x 20
30 10
=
10
30
=
1
3
= 0.3̄
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
(e) P(10 ≤ X ≤ 20) =
R 20
1
10 30−0 dx
=
P(X )
x 20
30 10
=
Area= 0.3̄
0.03̄
0
10
20
µ = 15
30
10
30
=
1
3
= 0.3̄
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Outline
1
The Uniform Distribution
2
The Normal Distribution
3
The Normal Approximation to the Binomial Distribution
4
The Normal Approximation to the Poisson Distribution
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The normal distribution, which we shall study in this section, is in
many ways the cornerstone of modern statistical theory.
Definition 4
A random variable X has a normal distribution and it is referred
to as a normal random variable if and only if its probability density
is given by
1 x−µ 2
1
n(x; µ, σ) = √ e − 2 ( σ ) for − ∞ < x < ∞
σ 2π
where σ > 0.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Characteristics of a Normal Distribution
It is bell-shaped and has a single peak at the center of the
distribution.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Characteristics of a Normal Distribution
It is bell-shaped and has a single peak at the center of the
distribution.
It is symmetrical about the mean.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Characteristics of a Normal Distribution
It is bell-shaped and has a single peak at the center of the
distribution.
It is symmetrical about the mean.
It is asymptotic: The curve gets closer and closer to the x-axis
but never actually touches it. To put it another way, the tails
of the curve extend indefinitely in both directions.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Characteristics of a Normal Distribution
It is bell-shaped and has a single peak at the center of the
distribution.
It is symmetrical about the mean.
It is asymptotic: The curve gets closer and closer to the x-axis
but never actually touches it. To put it another way, the tails
of the curve extend indefinitely in both directions.
The location of a normal distribution is determined by the
mean, µ, the dispersion or spread of the distribution is
determined by the standard deviation, σ.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Characteristics of a Normal Distribution
It is bell-shaped and has a single peak at the center of the
distribution.
It is symmetrical about the mean.
It is asymptotic: The curve gets closer and closer to the x-axis
but never actually touches it. To put it another way, the tails
of the curve extend indefinitely in both directions.
The location of a normal distribution is determined by the
mean, µ, the dispersion or spread of the distribution is
determined by the standard deviation, σ.
The arithmetic mean, median, and mode are equal.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Characteristics of a Normal Distribution
It is bell-shaped and has a single peak at the center of the
distribution.
It is symmetrical about the mean.
It is asymptotic: The curve gets closer and closer to the x-axis
but never actually touches it. To put it another way, the tails
of the curve extend indefinitely in both directions.
The location of a normal distribution is determined by the
mean, µ, the dispersion or spread of the distribution is
determined by the standard deviation, σ.
The arithmetic mean, median, and mode are equal.
The total area under the curve is 1.00; half the area under the
normal curve is to the right of this center point and the other
half to the left of it.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Characteristics of a Normal Distribution
It is bell-shaped and has a single peak at the center of the
distribution.
It is symmetrical about the mean.
It is asymptotic: The curve gets closer and closer to the x-axis
but never actually touches it. To put it another way, the tails
of the curve extend indefinitely in both directions.
The location of a normal distribution is determined by the
mean, µ, the dispersion or spread of the distribution is
determined by the standard deviation, σ.
The arithmetic mean, median, and mode are equal.
The total area under the curve is 1.00; half the area under the
normal curve is to the right of this center point and the other
half to the left of it.
Z ∞
Z ∞
1 x−µ 2
1
√ e − 2 ( σ ) dx = 1.
n(x; µ, σ)dx =
−∞ σ 2π
−∞
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The Normal Distribution - Graphically
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Equal Means and
Different Standard Deviations
Different Means and
Standard Deviations
Different Means and Equal Standard Deviations
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Theorem 5
The moment-generating function of the normal distribution is
given by
1 2 2
MX (t) = e µt+ 2 σ t .
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Since the normal distribution plays a basic role in statistics and its
density cannot be integrated directly, its areas have been tabulated
for the special case, where µ = 0 and σ = 1.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Since the normal distribution plays a basic role in statistics and its
density cannot be integrated directly, its areas have been tabulated
for the special case, where µ = 0 and σ = 1.
Definition 6
Let X be a random variable with mean µ∗ and standard deviation
σ ∗ (σ ∗ > 0). Then we can define an associated standard random
variable given by
X − µ∗
Z=
.
σ∗
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Since the normal distribution plays a basic role in statistics and its
density cannot be integrated directly, its areas have been tabulated
for the special case, where µ = 0 and σ = 1.
Definition 6
Let X be a random variable with mean µ∗ and standard deviation
σ ∗ (σ ∗ > 0). Then we can define an associated standard random
variable given by
X − µ∗
Z=
.
σ∗
µZ = E (Z ) = E
σZ2 = Var (Z ) = E
=
X − µ∗
σ∗
X−
σ∗
µ∗
=
1
1
E (X − µ∗ ) = ∗ [E (X ) −µ∗ ] = 0,
∗
σ
σ | {z }
µ∗
1
1
∗ 2
E [(X − µ∗ )2 ] = 1 ⇒ σZ = 1.
2 E [(X − µ ) ] =
∗
{z
}
σ
σ ∗2 |
∗2
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Definition 7
The normal distribution with µ = 0 and σ = 1 is referred to as the
standard normal distribution.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Definition 7
The normal distribution with µ = 0 and σ = 1 is referred to as the
standard normal distribution.
µ
The red curve is the standard normal distribution
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The entries in the table, represented by the shaded area of the figure, are
the values of
Z z
1 2
1
√ e − 2 x dx
2π
0
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The entries in the table, represented by the shaded area of the figure, are
the values of
Z z
1 2
1
√ e − 2 x dx
2π
0
that is the probabilities that a random variable having the standard
normal distribution will take on a value on the interval from 0 to z, for
z = 0.00, 0.01, 0.02, · · · , 3.08 and 3.09 and also z = 4.0, z = 5.0, and
z = 6.0.
R 1.5
0
1 2
√1 e − 2 x dx
2π
= 0.4332
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 8
Find the probabilities that a random variable having the standard
normal distribution will take on a value
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 8
Find the probabilities that a random variable having the standard
normal distribution will take on a value
(a) less that 1.72;
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 8
Find the probabilities that a random variable having the standard
normal distribution will take on a value
(a) less that 1.72;
(b) less than −0.88;
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 8
Find the probabilities that a random variable having the standard
normal distribution will take on a value
(a) less that 1.72;
(b) less than −0.88;
(c) between 1.3 and 1.75;
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 8
Find the probabilities that a random variable having the standard
normal distribution will take on a value
(a) less that 1.72;
(b) less than −0.88;
(c) between 1.3 and 1.75;
(d) between −0.25 and 0.45.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 8
Find the probabilities that a random variable having the standard
normal distribution will take on a value
(a) less that 1.72;
(b) less than −0.88;
(c) between 1.3 and 1.75;
(d) between −0.25 and 0.45.
Solution.
(a) P(Z < 1.72) = 0.5 + 0.4573 = 0.9573;
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 8
Find the probabilities that a random variable having the standard
normal distribution will take on a value
(a) less that 1.72;
(b) less than −0.88;
(c) between 1.3 and 1.75;
(d) between −0.25 and 0.45.
Solution.
(a) P(Z < 1.72) = 0.5 + 0.4573 = 0.9573;
(b) P(Z < −0.88) = 0.5 − 0.3106 = 0.1894;
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 8
Find the probabilities that a random variable having the standard
normal distribution will take on a value
(a) less that 1.72;
(b) less than −0.88;
(c) between 1.3 and 1.75;
(d) between −0.25 and 0.45.
Solution.
(a) P(Z < 1.72) = 0.5 + 0.4573 = 0.9573;
(b) P(Z < −0.88) = 0.5 − 0.3106 = 0.1894;
(c) P(1.3 < Z < 1.75) = 0.4599 − 0.4032 = 0.0567;
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 8
Find the probabilities that a random variable having the standard
normal distribution will take on a value
(a) less that 1.72;
(b) less than −0.88;
(c) between 1.3 and 1.75;
(d) between −0.25 and 0.45.
Solution.
(a) P(Z < 1.72) = 0.5 + 0.4573 = 0.9573;
(b) P(Z < −0.88) = 0.5 − 0.3106 = 0.1894;
(c) P(1.3 < Z < 1.75) = 0.4599 − 0.4032 = 0.0567;
(d) P(−0.25 < Z < 0.45) = 0.0987 + 0.1736 = 0.2727.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 9
With the reference to Standard Normal Distribution Table, find the
values of z that correspond to entries of
(a) 0.3512;
(b) 0.2533.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 9
With the reference to Standard Normal Distribution Table, find the
values of z that correspond to entries of
(a) 0.3512;
(b) 0.2533.
Solution.
(a) Since 0.3512 falls between 0.3508 and 0.3531, corresponding
to z = 1.04 and z = 1.05, and since 0.3512 is closer to 0.3508
than 0.3531, we choose z = 1.04.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 9
With the reference to Standard Normal Distribution Table, find the
values of z that correspond to entries of
(a) 0.3512;
(b) 0.2533.
Solution.
(a) Since 0.3512 falls between 0.3508 and 0.3531, corresponding
to z = 1.04 and z = 1.05, and since 0.3512 is closer to 0.3508
than 0.3531, we choose z = 1.04.
(b) Since 0.2533 falls midway between 0.2517 and 0.2549,
corresponding to z = 0.68 and z = 0.69, we choose
z = 0.685.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 10
Find z if the standard-normal-curve area
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 10
Find z if the standard-normal-curve area
(a) between 0 and z is 0.4726;
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 10
Find z if the standard-normal-curve area
(a) between 0 and z is 0.4726;
(b) to the left of z is 0.9869;
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 10
Find z if the standard-normal-curve area
(a) between 0 and z is 0.4726;
(b) to the left of z is 0.9869;
(c) to the right of z is 0.1314;
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 10
Find z if the standard-normal-curve area
(a) between 0 and z is 0.4726;
(b) to the left of z is 0.9869;
(c) to the right of z is 0.1314;
(d) between −z and z is 0.8502.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 10
Find z if the standard-normal-curve area
(a) between 0 and z is 0.4726;
(b) to the left of z is 0.9869;
(c) to the right of z is 0.1314;
(d) between −z and z is 0.8502.
Solution.
(a) z = ±1.92
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 10
Find z if the standard-normal-curve area
(a) between 0 and z is 0.4726;
(b) to the left of z is 0.9869;
(c) to the right of z is 0.1314;
(d) between −z and z is 0.8502.
Solution.
(a) z = ±1.92
(b) 0.9869 − 0.5 = 0.4869 ⇒ z = 2.22;
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 10
Find z if the standard-normal-curve area
(a) between 0 and z is 0.4726;
(b) to the left of z is 0.9869;
(c) to the right of z is 0.1314;
(d) between −z and z is 0.8502.
Solution.
(a) z = ±1.92
(b) 0.9869 − 0.5 = 0.4869 ⇒ z = 2.22;
(c) 0.5 − 0.1314 = 0.3686 ⇒ z = 1.12;
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 10
Find z if the standard-normal-curve area
(a) between 0 and z is 0.4726;
(b) to the left of z is 0.9869;
(c) to the right of z is 0.1314;
(d) between −z and z is 0.8502.
Solution.
(a) z = ±1.92
(b) 0.9869 − 0.5 = 0.4869 ⇒ z = 2.22;
(c) 0.5 − 0.1314 = 0.3686 ⇒ z = 1.12;
(d) 0.8502 ÷ 2 = 0.4251 ⇒ z = ±1.44.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
To determine probabilities related to random variables having
normal distribution other than the standard normal distribution, we
make use of the following theorem:
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
To determine probabilities related to random variables having
normal distribution other than the standard normal distribution, we
make use of the following theorem:
Theorem 11
If X has a normal distribution with mean µ and the standard
deviation σ, then
X −µ
Z=
σ
has the standard normal distribution.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
To determine probabilities related to random variables having
normal distribution other than the standard normal distribution, we
make use of the following theorem:
Theorem 11
If X has a normal distribution with mean µ and the standard
deviation σ, then
X −µ
Z=
σ
has the standard normal distribution.
Thus, to use the Standard Normal Distribution Table in connection
with any random variable having a normal distribution, we simply
perform the change of scale z = x−µ
σ .
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 12
Suppose that the amount of cosmic radiation to which a person is
exposed when flying by jet across the US is a random variable having a
normal distribution with mean of 4.35 mrem and a standard deviation of
0.59 mrem. What is the probability that a person will be exposed to
more that 5.20 mrem of cosmic radiation on such a flight?
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 12
Suppose that the amount of cosmic radiation to which a person is
exposed when flying by jet across the US is a random variable having a
normal distribution with mean of 4.35 mrem and a standard deviation of
0.59 mrem. What is the probability that a person will be exposed to
more that 5.20 mrem of cosmic radiation on such a flight?
Solution. Looking up the entry corresponding to
5.20 − 4.35
= 1.44
z=
0.59
in the Standard Normal Distribution Table and subtracting it from
0.5, we get
0.5 − 0.4251 = 0.0749.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 13
Suppose that during periods of transcendental meditation the
reduction of a person’s oxygen consumption is a random variable
having a normal distribution with µ = 37.6 cc per minute and
σ = 4.6 cc per minute. Find the probabilities that during a period
of transcendental meditation a person’s oxygen consumption will
be reduced by
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 13
Suppose that during periods of transcendental meditation the
reduction of a person’s oxygen consumption is a random variable
having a normal distribution with µ = 37.6 cc per minute and
σ = 4.6 cc per minute. Find the probabilities that during a period
of transcendental meditation a person’s oxygen consumption will
be reduced by
(a) at least 44.5 cc per minute;
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 13
Suppose that during periods of transcendental meditation the
reduction of a person’s oxygen consumption is a random variable
having a normal distribution with µ = 37.6 cc per minute and
σ = 4.6 cc per minute. Find the probabilities that during a period
of transcendental meditation a person’s oxygen consumption will
be reduced by
(a) at least 44.5 cc per minute;
(b) at most 35.0 cc per minute;
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 13
Suppose that during periods of transcendental meditation the
reduction of a person’s oxygen consumption is a random variable
having a normal distribution with µ = 37.6 cc per minute and
σ = 4.6 cc per minute. Find the probabilities that during a period
of transcendental meditation a person’s oxygen consumption will
be reduced by
(a) at least 44.5 cc per minute;
(b) at most 35.0 cc per minute;
(c) anywhere from 30.0 to 40.0 cc per minute.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Solution.
(a) at least 44.5 cc per minute: Looking up the entry
corresponding to
44.5 − 37.6
Z=
= 1.5
4.6
in the Standard Normal Distribution Table and subtracting it
from 0.5, we get
P(Z ≥ 1.5) = 0.5 − 0.4332 = 0.0668.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Solution.
(a) at least 44.5 cc per minute: Looking up the entry
corresponding to
44.5 − 37.6
Z=
= 1.5
4.6
in the Standard Normal Distribution Table and subtracting it
from 0.5, we get
P(Z ≥ 1.5) = 0.5 − 0.4332 = 0.0668.
(b) at most 35.0 cc per minute: Looking up the entry
corresponding to
35.0 − 37.6
Z=
= −0.565
4.6
in the Standard Normal Distribution Table as
0.2123+0.2157
= 0.2140 and subtracting it from 0.5, we get
2
P(Z ≤ −0.565) = 0.5 − 0.2140 = 0.2860.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
(c) anywhere from 30.0 to 40.0 cc per minute: Looking up the
entries corresponding to
Z1 =
30 − 37.6
= −1.6522 ≈ −1.65
4.6
and
40 − 37.6
= 0.5217 ≈ 0.52
4.6
in the Standard Normal Distribution Table, we obtain that
Z2 =
P(−1.65 ≤ Z ≤ 0.52) = 0.4505 + 0.1985 = 0.6490.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 14
A random variable has a normal distribution with σ = 10. If the
probability that the random variable will take on a value less than
82.5 is 0.8212, what is the probability that it will take on a value
greater than 58.3?
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 14
A random variable has a normal distribution with σ = 10. If the
probability that the random variable will take on a value less than
82.5 is 0.8212, what is the probability that it will take on a value
greater than 58.3?
Solution.
Total area: 0.8212
Area of rhs: 0.3212
0.92 =
82.5−µ
10
Since 0.8212 > 0.5, we see that z lies on the right hand side of the
mean.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 14
A random variable has a normal distribution with σ = 10. If the
probability that the random variable will take on a value less than
82.5 is 0.8212, what is the probability that it will take on a value
greater than 58.3?
Solution.
Total area: 0.8212
Area of rhs: 0.3212
0.92 =
82.5−µ
10
Since 0.8212 > 0.5, we see that z lies on the right hand side of the
mean. Then, 0.8212 − 0.5 = 0.3212 is the area between 0 and z.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 14
A random variable has a normal distribution with σ = 10. If the
probability that the random variable will take on a value less than
82.5 is 0.8212, what is the probability that it will take on a value
greater than 58.3?
Solution.
Total area: 0.8212
Area of rhs: 0.3212
0.92 =
82.5−µ
10
Since 0.8212 > 0.5, we see that z lies on the right hand side of the
mean. Then, 0.8212 − 0.5 = 0.3212 is the area between 0 and z.
By using the Standard Normal Distribution Table we obtain that
z = 0.92.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Therefore
0.92 =
82.5 − µ
⇒ µ = 73.3.
10
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Therefore
82.5 − µ
⇒ µ = 73.3.
10
Thus the probability that it will take on a value greater than 58.3 is
0.92 =
z=
58.3 − 73.3
= −1.5 ⇒
10
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Therefore
82.5 − µ
⇒ µ = 73.3.
10
Thus the probability that it will take on a value greater than 58.3 is
0.92 =
z=
58.3 − 73.3
= −1.5 ⇒
10
P(Z > −1.5) = 0.5 + 0.4332 = 0.9332.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Outline
1
The Uniform Distribution
2
The Normal Distribution
3
The Normal Approximation to the Binomial Distribution
4
The Normal Approximation to the Poisson Distribution
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
We have already seen that the Poisson distribution can be used to
approximate the binomial distribution for large values of n and
small values of θ provided that the correct conditions exist.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
We have already seen that the Poisson distribution can be used to
approximate the binomial distribution for large values of n and
small values of θ provided that the correct conditions exist. The
approximation is only of practical use if just a few terms of the
Poisson distribution need be calculated.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
We have already seen that the Poisson distribution can be used to
approximate the binomial distribution for large values of n and
small values of θ provided that the correct conditions exist. The
approximation is only of practical use if just a few terms of the
Poisson distribution need be calculated. In cases where many sometimes several hundred - terms need to be calculated the
arithmetic involved becomes very tedious indeed and we turn to
the normal distribution for help.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The normal distribution is sometimes introduced as a continuous
distribution that provides a close approximation to the binomial
distribution when n, the number of trials, is very large and θ, the
probability of success on an individual trials, is close to 1/2.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The normal distribution is sometimes introduced as a continuous
distribution that provides a close approximation to the binomial
distribution when n, the number of trials, is very large and θ, the
probability of success on an individual trials, is close to 1/2. The graph
of the binomial distribution where n = 50 and θ = 0.5 is shown in the
following figure.
Notice that it approximates a normal distribution. This suggests that a
binomial distribution can be approximated by a normal distribution as
long as the number of trials is relatively large.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Theorem 15
If X is a random variable having a binomial distribution with the
parameters n and θ, then the moment-generating function of
X − nθ
Z=p
nθ(1 − θ)
approaches that of the standard normal distribution when n → ∞.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Theorem 15
If X is a random variable having a binomial distribution with the
parameters n and θ, then the moment-generating function of
X − nθ
Z=p
nθ(1 − θ)
approaches that of the standard normal distribution when n → ∞.
Remark.
If X is a binomial random variable of n independent trials, each
with probability of success θ, and if
nθ > 5 and n(1 − θ) > 5
then the binomial random variable can be approximated by a
normal distribution with
p
µ = nθ and σ = nθ(1 − θ).
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The Continuity Correction
Consider an experiment where we toss a fair coin 12 times and observe
the number of heads.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The Continuity Correction
Consider an experiment where we toss a fair coin 12 times and observe
the number of heads. Suppose we want to compute the probability of
obtaining exactly 4 heads.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The Continuity Correction
Consider an experiment where we toss a fair coin 12 times and observe
the number of heads. Suppose we want to compute the probability of
obtaining exactly 4 heads. Whereas a discrete random variable can have
only a specified value (such as 4), a continuous random variable used to
approximate it could take on any values within an interval around that
specified value,
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The Continuity Correction
Consider an experiment where we toss a fair coin 12 times and observe
the number of heads. Suppose we want to compute the probability of
obtaining exactly 4 heads. Whereas a discrete random variable can have
only a specified value (such as 4), a continuous random variable used to
approximate it could take on any values within an interval around that
specified value, as demonstrated in this figure:
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The Continuity Correction
Consider an experiment where we toss a fair coin 12 times and observe
the number of heads. Suppose we want to compute the probability of
obtaining exactly 4 heads. Whereas a discrete random variable can have
only a specified value (such as 4), a continuous random variable used to
approximate it could take on any values within an interval around that
specified value, as demonstrated in this figure:
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The continuity correction requires adding or subtracting 0.5 from
the value or values of the discrete random variable X as needed.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The continuity correction requires adding or subtracting 0.5 from
the value or values of the discrete random variable X as needed.
Hence to use the normal distribution to approximate the
probability of obtaining exactly 4 heads (i.e., X = 4), we would
find the area under the normal curve from X = 3.5 to X = 4.5, the
lower and upper boundaries of 4.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Moreover, to determine the approximate probability of observing at
least 4 heads, we would find the area under the normal curve from
X = 3.5 and above since, on a continuum, 3.5 is the lower
boundary of X .
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Moreover, to determine the approximate probability of observing at
least 4 heads, we would find the area under the normal curve from
X = 3.5 and above since, on a continuum, 3.5 is the lower
boundary of X . Similarly, to determine the approximate probability
of observing at most 4 heads, we would find the area under the
normal curve from X = 4.5 and below since, on a continuum, 4.5
is the upper boundary of X .
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Now consider the binomial distribution in particular.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Now consider the binomial distribution in particular. Let X be the
number of heads in 12 flips of a fair coin.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Now consider the binomial distribution in particular. Let X be the
number of heads in 12 flips of a fair coin. We know that the
probability of observing exactly x heads in 12 flips is
12
0.5x 0.512−x , x = 0, 1, 2 · · · , 12.
P(X = x) = b(x; 12, 0.5) =
x
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Now consider the binomial distribution in particular. Let X be the
number of heads in 12 flips of a fair coin. We know that the
probability of observing exactly x heads in 12 flips is
12
0.5x 0.512−x , x = 0, 1, 2 · · · , 12.
P(X = x) = b(x; 12, 0.5) =
x
We also know that the mean of this binomial distribution is given
by
µ = nθ = 12(0.5) = 6,
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Now consider the binomial distribution in particular. Let X be the
number of heads in 12 flips of a fair coin. We know that the
probability of observing exactly x heads in 12 flips is
12
0.5x 0.512−x , x = 0, 1, 2 · · · , 12.
P(X = x) = b(x; 12, 0.5) =
x
We also know that the mean of this binomial distribution is given
by
µ = nθ = 12(0.5) = 6,
We also know that the standard deviation of this binomial
distribution is given by
p
p
σ = nθ(1 − θ) = 12(0.5)(0.5) = 1.732.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Now consider the binomial distribution in particular. Let X be the
number of heads in 12 flips of a fair coin. We know that the
probability of observing exactly x heads in 12 flips is
12
0.5x 0.512−x , x = 0, 1, 2 · · · , 12.
P(X = x) = b(x; 12, 0.5) =
x
We also know that the mean of this binomial distribution is given
by
µ = nθ = 12(0.5) = 6,
We also know that the standard deviation of this binomial
distribution is given by
p
p
σ = nθ(1 − θ) = 12(0.5)(0.5) = 1.732.
Say we are interested in the probability of observing between 3 and
5 heads, inclusive;
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Now consider the binomial distribution in particular. Let X be the
number of heads in 12 flips of a fair coin. We know that the
probability of observing exactly x heads in 12 flips is
12
0.5x 0.512−x , x = 0, 1, 2 · · · , 12.
P(X = x) = b(x; 12, 0.5) =
x
We also know that the mean of this binomial distribution is given
by
µ = nθ = 12(0.5) = 6,
We also know that the standard deviation of this binomial
distribution is given by
p
p
σ = nθ(1 − θ) = 12(0.5)(0.5) = 1.732.
Say we are interested in the probability of observing between 3 and
5 heads, inclusive; that is, P(3 ≤ X ≤ 5).
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Now consider the binomial distribution in particular. Let X be the
number of heads in 12 flips of a fair coin. We know that the
probability of observing exactly x heads in 12 flips is
12
0.5x 0.512−x , x = 0, 1, 2 · · · , 12.
P(X = x) = b(x; 12, 0.5) =
x
We also know that the mean of this binomial distribution is given
by
µ = nθ = 12(0.5) = 6,
We also know that the standard deviation of this binomial
distribution is given by
p
p
σ = nθ(1 − θ) = 12(0.5)(0.5) = 1.732.
Say we are interested in the probability of observing between 3 and
5 heads, inclusive; that is, P(3 ≤ X ≤ 5). We can calculate this
exactly, of course:
P(3 ≤ X ≤ 5) = P(X = 3) + P(X = 4) + P(X = 5)
= 0.05371 + 0.12085 + 0.19336 = 0.36792.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
What about the normal approximation to this value?
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
What about the normal approximation to this value? First, we
should check whether the normal approximation is likely to work
very well in this case.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
What about the normal approximation to this value? First, we
should check whether the normal approximation is likely to work
very well in this case. A useful rule of thumb is that the normal
approximation should work well enough if both nθ and n(1 − θ) are
greater than 5.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
What about the normal approximation to this value? First, we
should check whether the normal approximation is likely to work
very well in this case. A useful rule of thumb is that the normal
approximation should work well enough if both nθ and n(1 − θ) are
greater than 5. For this example, both equal 12(0.5) = 6, so we
are about at the limit of usefulness of the approximation.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
What about the normal approximation to this value? First, we
should check whether the normal approximation is likely to work
very well in this case. A useful rule of thumb is that the normal
approximation should work well enough if both nθ and n(1 − θ) are
greater than 5. For this example, both equal 12(0.5) = 6, so we
are about at the limit of usefulness of the approximation.
Back to the question at hand. Since X is a binomial random
variable, the following statement (based on the continuity
correction) is exactly correct:
P(3 ≤ X ≤ 5) = P(2.5 < X < 5.5).
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
What about the normal approximation to this value? First, we
should check whether the normal approximation is likely to work
very well in this case. A useful rule of thumb is that the normal
approximation should work well enough if both nθ and n(1 − θ) are
greater than 5. For this example, both equal 12(0.5) = 6, so we
are about at the limit of usefulness of the approximation.
Back to the question at hand. Since X is a binomial random
variable, the following statement (based on the continuity
correction) is exactly correct:
P(3 ≤ X ≤ 5) = P(2.5 < X < 5.5).
Note that this statement is not an approximation - it is exactly
correct!
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
What about the normal approximation to this value? First, we
should check whether the normal approximation is likely to work
very well in this case. A useful rule of thumb is that the normal
approximation should work well enough if both nθ and n(1 − θ) are
greater than 5. For this example, both equal 12(0.5) = 6, so we
are about at the limit of usefulness of the approximation.
Back to the question at hand. Since X is a binomial random
variable, the following statement (based on the continuity
correction) is exactly correct:
P(3 ≤ X ≤ 5) = P(2.5 < X < 5.5).
Note that this statement is not an approximation - it is exactly
correct! The reason for this is that we are adding the events
2.5 < X < 3 and 5 < X < 5.5 to get from the left side of the
equation to the right side of the equation, but for the binomial
random variable, these events have probability zero.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The continuity correction is not where the approximation comes
in; that comes when we approximate X using a normal distribution
with mean µ = 6 and standard deviation σ = 1.732:
P(3 ≤ X ≤ 5) = P(2.5 < X < 5.5)
5.5 − 6
2.5 − 6
<Z <
≈P
1.732
1.732
= P(−2.02 < Z < −0.29)
= 0.4783 − 0.1141 = 0.3642.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The continuity correction is not where the approximation comes
in; that comes when we approximate X using a normal distribution
with mean µ = 6 and standard deviation σ = 1.732:
P(3 ≤ X ≤ 5) = P(2.5 < X < 5.5)
5.5 − 6
2.5 − 6
<Z <
≈P
1.732
1.732
= P(−2.02 < Z < −0.29)
= 0.4783 − 0.1141 = 0.3642.
Note that the approximation is only off by about 1%, which is
pretty good for such a small sample size.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 16
Suppose that a sample of n = 1, 600 tires of the same type are
obtained at random from an ongoing production process in which
8% of all such tires produced are defective. What is the probability
that in such a sample not more than 150 tires will be defective?
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 16
Suppose that a sample of n = 1, 600 tires of the same type are
obtained at random from an ongoing production process in which
8% of all such tires produced are defective. What is the probability
that in such a sample not more than 150 tires will be defective?
Solution. Let X be the number of selected defective items.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 16
Suppose that a sample of n = 1, 600 tires of the same type are
obtained at random from an ongoing production process in which
8% of all such tires produced are defective. What is the probability
that in such a sample not more than 150 tires will be defective?
Solution. Let X be the number of selected defective items. The
mean and the standard deviation are
µ = nθ = 1, 600(0.08) = 128(> 5)(n(1−θ) = 1, 600(0.92) = 1, 472 > 0),
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 16
Suppose that a sample of n = 1, 600 tires of the same type are
obtained at random from an ongoing production process in which
8% of all such tires produced are defective. What is the probability
that in such a sample not more than 150 tires will be defective?
Solution. Let X be the number of selected defective items. The
mean and the standard deviation are
µ = nθ = 1, 600(0.08) = 128(> 5)(n(1−θ) = 1, 600(0.92) = 1, 472 > 0),
σ=
p
p
nθ(1 − θ) = 1, 600(0.08)(0.92) = 10.85.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 16
Suppose that a sample of n = 1, 600 tires of the same type are
obtained at random from an ongoing production process in which
8% of all such tires produced are defective. What is the probability
that in such a sample not more than 150 tires will be defective?
Solution. Let X be the number of selected defective items. The
mean and the standard deviation are
µ = nθ = 1, 600(0.08) = 128(> 5)(n(1−θ) = 1, 600(0.92) = 1, 472 > 0),
σ=
p
p
nθ(1 − θ) = 1, 600(0.08)(0.92) = 10.85.
The probability calculation is thus
P(X ≤ 150) = P(X < 150.5)
150.5 − 128
≈P Z <
= P(Z < 2.07)
10.85
= 0.5 + 0.4808 = 0.9808.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 17
Based on past experience, 7% of all luncheon tickets are in error. If
a random sample of 400 tickets is selected, what is the
approximate probability that
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 17
Based on past experience, 7% of all luncheon tickets are in error. If
a random sample of 400 tickets is selected, what is the
approximate probability that
(a) exactly 25 are in error?
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 17
Based on past experience, 7% of all luncheon tickets are in error. If
a random sample of 400 tickets is selected, what is the
approximate probability that
(a) exactly 25 are in error?
(b) at most 25 are in error?
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 17
Based on past experience, 7% of all luncheon tickets are in error. If
a random sample of 400 tickets is selected, what is the
approximate probability that
(a) exactly 25 are in error?
(b) at most 25 are in error?
(c) between 20 and 25 (inclusive) are in error?
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 17
Based on past experience, 7% of all luncheon tickets are in error. If
a random sample of 400 tickets is selected, what is the
approximate probability that
(a) exactly 25 are in error?
(b) at most 25 are in error?
(c) between 20 and 25 (inclusive) are in error?
Solution. Let X be the number of selected tickets in error.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 17
Based on past experience, 7% of all luncheon tickets are in error. If
a random sample of 400 tickets is selected, what is the
approximate probability that
(a) exactly 25 are in error?
(b) at most 25 are in error?
(c) between 20 and 25 (inclusive) are in error?
Solution. Let X be the number of selected tickets in error. The
mean and the standard deviation are
µ = nθ = 400(0.07) = 28,
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 17
Based on past experience, 7% of all luncheon tickets are in error. If
a random sample of 400 tickets is selected, what is the
approximate probability that
(a) exactly 25 are in error?
(b) at most 25 are in error?
(c) between 20 and 25 (inclusive) are in error?
Solution. Let X be the number of selected tickets in error. The
mean and the standard deviation are
µ = nθ = 400(0.07) = 28,
σ=
p
p
nθ(1 − θ) = 400(0.07)(0.93) = 5.103.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 17
Based on past experience, 7% of all luncheon tickets are in error. If
a random sample of 400 tickets is selected, what is the
approximate probability that
(a) exactly 25 are in error?
(b) at most 25 are in error?
(c) between 20 and 25 (inclusive) are in error?
Solution. Let X be the number of selected tickets in error. The
mean and the standard deviation are
µ = nθ = 400(0.07) = 28,
σ=
p
p
nθ(1 − θ) = 400(0.07)(0.93) = 5.103.
The probability calculations are thus
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
(a) exactly 25 are in error?
P(X = 25) = P(24.5 < X < 25.5)
24.5 − 28
25.5 − 28
≈P
<Z <
5.103
5.103
= P(−0.69 < Z < −0.49)
= 0.2549 − 0.1879 = 0.067
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
(a) exactly 25 are in error?
P(X = 25) = P(24.5 < X < 25.5)
24.5 − 28
25.5 − 28
≈P
<Z <
5.103
5.103
= P(−0.69 < Z < −0.49)
= 0.2549 − 0.1879 = 0.067
(b) at most 25 are in error?
P(X ≤ 25) = P(X < 25.5)
25.5 − 28
≈P Z <
5.103
= P(Z < −0.49) = 0.5 − 0.1879 = 0.3121.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
(c) between 20 and 25 (inclusive) are in error?
P(20 ≤ X ≤ 25) = P(19.5 < X < 25.5)
25.5 − 28
19.5 − 28
<Z <
≈P
5.103
5.103
= P(−1.67 < Z < −0.49)
= 0.4525 − 0.1879 = 0.2646.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 18
An airline knows that over the long run, 90% of passengers who
reserve seats will show up for their fights. On a particular fight
with 300 seats, the airline accepts 324 reservations. Use the
normal approximation to the binomial distribution to find the
chance that the fight will be overbooked?
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 18
An airline knows that over the long run, 90% of passengers who
reserve seats will show up for their fights. On a particular fight
with 300 seats, the airline accepts 324 reservations. Use the
normal approximation to the binomial distribution to find the
chance that the fight will be overbooked?
Solution. Let X be the numberpof passengers. Since
µ = 324(0.9) = 291.6 and σ = 324(0.9)(0.1) = 5.4, then
P(X ≥ 301) = P(X > 300.5)
300.5 − 291.6
≈P Z >
5.4
= P(Z > 1.65)
= 0.5 − 0.4505
= 0.0495
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 19
IQ scores of people around the world are normally distributed, with
a mean of 100 and a standard deviation of 15. A genius is
someone with an IQ greater than or equal to 140. What percent of
the population is considered genius?
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 19
IQ scores of people around the world are normally distributed, with
a mean of 100 and a standard deviation of 15. A genius is
someone with an IQ greater than or equal to 140. What percent of
the population is considered genius?
Solution.
P(X ≥ 140) = P(X > 139.5)
139.5 − 100
≈P Z >
15
= P(Z > 2.63)
= 0.5 − 0.4957
= 0.0043
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Outline
1
The Uniform Distribution
2
The Normal Distribution
3
The Normal Approximation to the Binomial Distribution
4
The Normal Approximation to the Poisson Distribution
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The normal distribution can also be used to approximate the
Poisson distribution for large values of λ.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The normal distribution can also be used to approximate the
Poisson distribution for large values of λ. The Poisson distribution
is also a discrete random variable, whereas the normal distribution
is continuous.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The normal distribution can also be used to approximate the
Poisson distribution for large values of λ. The Poisson distribution
is also a discrete random variable, whereas the normal distribution
is continuous. Therefore, we need to take this into account when
we are using the normal distribution to approximate a Poisson
using a continuity correction.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The normal distribution can also be used to approximate the
Poisson distribution for large values of λ. The Poisson distribution
is also a discrete random variable, whereas the normal distribution
is continuous. Therefore, we need to take this into account when
we are using the normal distribution to approximate a Poisson
using a continuity correction.
If X is a Poisson random variable with µ = λ and σ 2 = λ,
X −λ
Z= √
λ
is approximately a standard normal variable.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The normal distribution can also be used to approximate the
Poisson distribution for large values of λ. The Poisson distribution
is also a discrete random variable, whereas the normal distribution
is continuous. Therefore, we need to take this into account when
we are using the normal distribution to approximate a Poisson
using a continuity correction.
If X is a Poisson random variable with µ = λ and σ 2 = λ,
X −λ
Z= √
λ
is approximately a standard normal variable. The same continuity
correction used for the binomial distribution can also be applied.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
The normal distribution can also be used to approximate the
Poisson distribution for large values of λ. The Poisson distribution
is also a discrete random variable, whereas the normal distribution
is continuous. Therefore, we need to take this into account when
we are using the normal distribution to approximate a Poisson
using a continuity correction.
If X is a Poisson random variable with µ = λ and σ 2 = λ,
X −λ
Z= √
λ
is approximately a standard normal variable. The same continuity
correction used for the binomial distribution can also be applied.
The approximation is good for
λ > 5.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 20
Assume that the number of asbestos particles in a squared meter
of dust on a surface fallows a Poisson distribution with a mean of
1000. If a squared meter of dust is analyzed, what is the
probability that 950 or fewer particles are found?
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 20
Assume that the number of asbestos particles in a squared meter
of dust on a surface fallows a Poisson distribution with a mean of
1000. If a squared meter of dust is analyzed, what is the
probability that 950 or fewer particles are found?
Solution. This probability can be expressed exactly as
P(X ≤ 950) =
950 −1000
X
e
1000x
x=0
x!
.
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Example 20
Assume that the number of asbestos particles in a squared meter
of dust on a surface fallows a Poisson distribution with a mean of
1000. If a squared meter of dust is analyzed, what is the
probability that 950 or fewer particles are found?
Solution. This probability can be expressed exactly as
P(X ≤ 950) =
950 −1000
X
e
1000x
x=0
x!
.
The computational difficulty is clear. The probability can be
approximated as
P(X ≤ 950) = P(X ≤ 950.5)
950.5 − 1000
√
≈P Z ≤
1000
= P(Z ≤ −1.565) = 0.5 −
0.4406 + 0.4418
= 0.0588.
2
The Uniform Distribution The Normal Distribution The Normal Approximation to the Binomial Distribution The Normal Approxim
Thank You!!!