Download Applications of Normal Distribution

Chapter 12
Elementary Statistics
Applications
of
Normal Distribution
Chapter 12
Elementary Statistics
Applications of Normal Distribution
◮
Blood pressure of adults at certain age
◮
Length of time of a football game
◮
Speed of cars on a certain road
◮
Price of similar vehicles
◮
Monthly income
◮
And · · ·
Chapter 11
Elementary Statistics
Chapter 12
Elementary Statistics
Example:
The filling process dispenses cookies with weight that follows the
normal distribution with mean 510 grams and standard deviation 5
grams. A bag of cookies is underweight if it weighs less than 500
grams and is overweight if it weighs more than 520 grams.
◮
What is the probability that a randomly selected bag is
underweight?
◮
What is the probability that a randomly selected bag is
overweight?
◮
What is the probability that a randomly selected bag weighs
between 505 and 520 grams?
Solution:
We have a normal probability distribution with µ = 510, and
σ = 5.
Chapter 12
Elementary Statistics
Solution Continued:
◮
What is the probability that a randomly selected bag is
underweight? ⇒ P(x < 500)
P(x < 500)
500
|
µ = 510
σ=5
Now we can use the normal distribution chart or different
technology to compute the area which is the probability that
we are looking for. In this case, P(x < 500) = 0.023.
Chapter 12
Elementary Statistics
Solution Continued:
◮
What is the probability that a randomly selected bag is
overweight? ⇒ P(x > 520)
P(x > 520)
|
µ = 510
520
σ=5
Now we can use the normal distribution chart or different
technology to compute the area which is the probability that
we are looking for. In this case, P(x > 520) = 0.023.
Chapter 12
Elementary Statistics
Solution Continued:
◮
What is the probability that a randomly selected bag weighs
between 505 and 520 grams? ⇒ P(505 < x < 520)
P(505 < x < 520)
505
|
µ = 510
520
σ=5
Now we can use the normal distribution chart or different
technology to compute the area which is the probability that
we are looking for. In this case, P(505 < x < 520) = 0.8186.
Chapter 11
Elementary Statistics
Chapter 12
Elementary Statistics
Example:
The data from traffic analysts INRIX suggests that at the
northbound 405, between the 10 and the 101 Freeways, over a
two-week period, between 4 and 7 pm, a total distance of 10 miles,
they found that traversing that stretch of the highway took 35
minutes, which suggests average speeds of 16 miles per hour. Let’s
suppose that speed of cars on that segment of the freeway during
those time follow a normal distribution with standard deviation 2
miles per hour.
◮
What is the probability that a randomly selected car has a
speed between 15 and 20 miles per hour?
◮
Find the speed of cars that separates the top 5% from the rest
of cars.
Solution:
We have a normal probability distribution with µ = 16, and σ = 2.
Chapter 12
Elementary Statistics
Solution Continued:
◮
What is the probability that a randomly selected car has a
speed between 15 and 20 miles per hour?⇒ P(15 < x < 20)
P(15 < x < 20)
15
|
µ = 16
20
σ=2
Now we can use the normal distribution chart or different
technology to compute the area which is the probability that
we are looking for. In this case, P(15 < x < 20) = 0.6687.
Chapter 12
Elementary Statistics
Solution Continued:
Find the speed of cars that separates the top 5% from the rest
of cars. ⇒ x = P95
0.05
0.95
|
µ = 16
P95
σ=2
Using any method, we get x = P95 ≈ 19.3, that is
P(x < 19.3) = 0.95 and P(x > 19.3) = 0.05. So 5% of the
cars have speed above 19.3 miles per hour.
Chapter 12
Elementary Statistics
The average teacher salary in California seems to always be higher
than the national average. In 2009, the average teacher salary in
California was $60,583, considerably higher than the national
average of $49,720.
Chapter 12
Elementary Statistics
Example:
Suppose the monthly salary of teachers with no more than 5 years
teaching experience in California follows a normal distribution with
the mean of $6250 and standard deviation of $320.
◮
What is the probability that a randomly selected teacher with
no more than 5 years teaching experience in California makes
less than $6000 or more than $6500 per month?
◮
Find a range of salaries, rounded to to the nearest ten, that
separates the middle 80% from the rest.
Solution:
We have a normal probability distribution with µ = 6250, and
σ = 320.
Chapter 12
Elementary Statistics
Solution Continued:
◮
What is the probability that a randomly selected teacher with
no more than 5 years teaching experience in California makes
less than $6000 or more than $6500 per month?
⇒ P(x < 6000 or x > 6500)
P(x > 6500)
P(x < 6000)
6000
|
µ = 6250
6500
σ = 320
Since the total area under the normal curve is equal to 1, then
P(x < 6000 or x > 6500) = 1 − P(6000 < x < 6500).
Chapter 12
Elementary Statistics
Solution Continued:
Using any method, we get P(6000 < x < 6500) = 0.5653,
therefore P(x < 6000 or x > 6500) = 1 − 0.5653 = 0.4347
Find a range of salaries, rounded to to the nearest ten, that
separates the middle 80% from the rest.
⇒ P(x1 < x < x2 ) = .8
10%
10%
Middle 80%
x1
|
µ = 6250
x2
σ = 320
Using any method, we get x1 = P10 ≈ 5840 , and
x2 = P90 ≈ 6660 .
Chapter 12
Elementary Statistics
Estimating Binomial Distribution
by
Using Normal Distribution
Whenever n · p ≥ 5 & n · q ≥ 5
Binomial Distribution
Normal Distribution
P(x = a)
P(a − 0.5 < x < a + 0.5)
P(x ≤ a)
P(x < a + 0.5)
P(x ≥ a)
P(x > a − 0.5)
P(a ≤ x ≤ b)
P(a − 0.5 < x < b + 0.5)
Chapter 12
Elementary Statistics
Example:
Consider a binomial probability distribution with n = 100, p = 0.5,
µ = 50, σ = 5, and x is the number of successes.
◮
Show that n · p ≥ 5 and n · q ≥ 5.
◮
Find P(x = 48)
◮
Estimate P(x = 48) by using normal distribution.
Solution:
Since p = 0.5, then q = 1 − p = 1 − 0.5 = 0.5,
◮
Show that n · p ≥ 5 and n · q ≥ 5.
⇒ n · p = 100 · 0.5 = 50 ≥ 5 and n · q = 100 · 0.5 = 50 ≥ 5
◮
Find P(x = 48)
⇒ P(x = 48) = binompdf(100, 0.5, 48) = 0.0735
Chapter 12
Elementary Statistics
Solution Continued:
◮
Estimate P(x = 48) by using normal distribution.
⇒ P(x = 48) ≈ P(47.5 < x < 48.5)
P(47.5 < x < 48.5) = normaldf(47.5, 48.5, 50, 5) = 0.0736
Example:
Consider a binomial probability distribution with n = 180, p = .75,
and x is the number of successes.
◮
◮
Show that n · p ≥ 5 and n · q ≥ 5.
Find P(x ≤ 145)
◮
Find µ
◮
Find σ
◮
Estimate P(x ≤ 145) by using normal distribution.
Chapter 12
Elementary Statistics
Solution:
Since p = 0.75, then q = 1 − p = 1 − 0.75 = 0.25,
◮
Show that n · p ≥ 5 and n · q ≥ 5.
⇒ n · p = 180 · 0.75 = 135 ≥ 5 and n · q = 180 · 0.25 = 45 ≥ 5
◮
Find P(x ≤ 145)
⇒ P(x ≤ 145) = binomcdf(180, 0.75, 145) = 0.9675
◮
Find µ
⇒ µ = n · p = 180 · 0.75 = 135
√
√
Find σ ⇒ σ = n · p · q = 180 · 0.75 · 0.25 = 5.809
◮
◮
Estimate P(x ≤ 145) by using normal distribution.
⇒ P(x ≤ 145) ≈ P(x < 145.5)
P(x < 145.5) = normaldf(−E 99, 145.5, 135, 5.809) = 0.965
Chapter 12
Elementary Statistics
Example:
A multiple-choice exam has several hundred questions, each with 4
possible answers of which only 1 is the correct answer.
Suppose we randomly select 80 questions, and x is the number of
correct answers by a student whose selections of answers is by
sheer guesswork with no knowledge of the subject that is being
tested on.
◮
Identify n, p and q.
◮
Show that n · p ≥ 5 and n · q ≥ 5.
◮
Find µ.
◮
Find σ.
◮
Use normal distribution to estimate the probability that sheer
guesswork yields from 10 to 15 correct answers.
Chapter 12
Elementary Statistics
Solution:
◮
◮
◮
◮
◮
1
Identify n, p and q. ⇒ n = 80, p = = 0.25, and
4
3
q = = 0.75
4
Show that n · p ≥ 5 and n · q ≥ 5.
⇒ n · p = 80 · 0.25 = 20 ≥ 5 and n · q = 80 · 0.75 = 60 ≥ 5
Find Find µ.
⇒ µ = n · p = 80 · 0.25 = 20
√
√
Find σ ⇒ σ = n · p · q = 80 · 0.25 · 0.75 = 3.873
Use normal distribution to estimate the probability that sheer
guesswork yields from 10 to 15 correct answers.
⇒ P(10 ≤ x ≤ 15) ≈ P(9.5 < x < 15.5)
P(9.5 < x < 15.5) = normaldf(9.5, 15.5, 20, 3.873) = 0.1193
Chapter 12
Elementary Statistics
What is a Sampling Distribution?
It is a probability distribution of a statistic such as sample mean x̄.
What is the Central Limit Theorem?
It is the conclusion of the sampling distribution of x̄ from any
population with mean µ and variance σ 2 when random samples of
size n are drawn from.
The sampling distribution of x̄
◮
is approximately normally distributed
◮
with mean µx̄ = µ,
σ2
σ
variance σx̄2 =
, and standard deviation σx̄ = √ .
n
n
◮
Chapter 12
Elementary Statistics
Example:
Consider a discrete population consisting of values 2, 4, 6, 8 and 10.
◮
Find µ and σ 2 .
◮
List all possible samples of size 2 with replacement.
◮
Find the mean of each samples.
◮
Construct a table that contains the mean of each samples and
the probability of each mean.
◮
Use the discrete probability distribution to compute the mean
of x̄, that is µx̄ .
◮
Use the discrete probability distribution to compute the
variance of x̄, that is σx̄2 .
◮
Use these results to verify the conclusion of the Central Limit
Theorem.
Chapter 12
Elementary Statistics
Solution:
◮
◮
Find µ and σ 2 . ⇒ µ = 6, and σ 2 = 8
List all possible samples of size 2 with replacement.
2,2
2,4
2,6
2,8
2,10
◮
4,2
4,4
4,6
4,8
4,10
6,2
6,4
6,6
6,8
6,10
8,2
8,4
8,6
8,8
8,10
Find the mean of each samples.
2
3
4
5
6
3
4
5
6
7
4
5
6
7
8
5
6
7
8
9
6
7
8
9
10
10,2
10,4
10,6
10,8
10,10
Chapter 12
Elementary Statistics
Solution Continued:
◮
◮
◮
Construct a table that contains the mean of each samples and
the probability of each mean.
x̄
2
3
4
5
6
7
8
9
10
P(x̄)
1
25
2
25
3
25
4
25
5
25
4
25
3
25
2
25
1
25
Use the discrete probability distribution to compute the mean
of x̄, that is µx̄ .
⇒ µx̄ = 6
Use the discrete probability distribution to compute the
variance of x̄, that is σx̄2 .
⇒ σx̄2 = 4
Chapter 12
Elementary Statistics
Solution Continued:
◮
Use these results to verify the conclusion of the Central Limit
Theorem.
8
σ2
= = 4.
We can verify that µx̄ = µ = 8, and σx̄2 =
n
2
Example:
Use sampling distribution of x̄ when samples of size 16 are selected
at random from a normally distributed population with mean 375
and variance 100.
◮
Find µx̄ .
◮
Find σx̄2 .
◮
Find σx̄ .
Chapter 12
Elementary Statistics
Solution:
Using the Central Limit Theorem,
◮
◮
◮
Find µx̄ . ⇒ µx̄ = µ = 375
σ2
100
=
= 6.25
n
16
q
√
Find σx̄ . ⇒ σx̄ = σx̄2 = 6.25 = 2.5
Find σx̄2 . ⇒ σx̄2 =
Example:
Use sampling distribution of x̄ when samples of size 10 are selected
at random from a normally distributed population with mean 82
and standard deviation 7.5.
◮
Find µx̄ .
◮
Find σx̄ .
Chapter 12
Elementary Statistics
Solution:
Using the Central Limit Theorem,
◮
◮
Find µx̄ . ⇒ µx̄ = µ = 82
σ
7.5
Find σx̄ . ⇒ σx̄ = √ = √ ≈ 2.372 .
n
10
Z score & x̄ Sampling Distribution
x −µ
, now we can replace x with x̄, µ with µx̄ ,
σ
x̄ − µ
x̄ − µx̄
to get z = σ .
σ with σx̄ , and reduce z =
σx̄
√
n
we know that z =
Chapter 12
Elementary Statistics
Example:
Use sampling distribution of x̄ when samples of size 36 are selected
at random from a normally distributed population with mean 6250
and standard deviation 275.
◮
Find the z score for x̄ = 6450.
◮
Find the z score for x̄ = 6200.
Solution:
Using the formula z =
◮
◮
x̄ − µ
σ ,
√
n
6450 − 6250
≈ 4.364
275
√
36
6200 − 6250
≈ −1.091
Find the z score for x̄ = 5820. ⇒ z =
275
√
36
Find the z score for x̄ = 6450. ⇒ z =
Chapter 12
Elementary Statistics
Example:
The average life of a certain blender is 5.1 years with a standard
deviation of 1.2 years. Assume the lives of these blenders are
normally distributed.
◮
Find the probability that a mean life of a random sample of 9
such blenders fall between 4.5 and 5.5 years.
◮
Find the value of x̄ that separates the top 10% from the rest
of the means computed from random samples of size 9.
Solution:
We have a normal probability distribution with µ = 5.1, σ = 1.2,
and random sample of size 9. We can use the central limit theorem
σ
1.2
to compute µx̄ = µ = 5.1 and σx̄ = √ = √ = 0.4
n
9
Chapter 12
Elementary Statistics
Solution Continued:
◮
Find the probability that that a mean life of a random sample
of 9 such blenders fall between 4.5 and 5.5 years.
⇒ P(4.5 < x̄ < 5.5)
4.5
|
µx̄ = 5.1
σx̄ = 0.4
5.5
P(4.5 < x̄ < 5.5) = normaldf(4.5, 5.5, 5.1, 0.4) = 0.7745
Chapter 12
Elementary Statistics
Solution Continued:
◮
Find the value of x̄ that separates the top 10% from the rest
of the means computed from random samples of size 9.
⇒ P(x̄ > k) = 0.1
90%
10%
|
µx̄ = 5.1
σx̄ = 0.4
k
x̄ = k = P90 = invNorm(0.9, 5.1, 0.4) ≈ 5.6
Chapter 12
Elementary Statistics
Example:
Suppose the hourly wages of all workers in a manufacturer
company have a normal distribution with a mean of $15.50 and a
standard deviation of $2.75. If we randomly select a sample of 10
workers from this company, find the probability that their mean
hourly wages is
◮
less than $14.25.
◮
more than $16.50.
Solution:
We have a normal probability distribution with µ = 15.50,
σ = 2.75, and random sample of size 10. We can use the central
limit theorem to compute µx̄ = µ = 15.50 and
σ
2.75
σx̄ = √ = √ ≈ 0.87
n
10
Chapter 12
Elementary Statistics
Solution Continued:
◮
less than $14.25.
⇒ P(x̄ < 14.25)
14.25
|
µx̄ = 15.50
σx̄ = 0.87
P(x̄ < 14.25) = normaldf(−E 99, 14.25, 15.5, 0.87) = 0.0754
Chapter 12
Elementary Statistics
Solution Continued:
◮
more than $16.50.
⇒ P(x̄ > 16.50)
P(x̄ > 16.50)
|
µx̄ = 15.50
σx̄ = 0.87
16.50
P(x̄ = 16.50) = normalcdf(16.50, E 99, 15.50, 0.87) ≈ 0.125
Chapter 11
Elementary Statistics