Download S2.3 Continuous distributions

A2-Level Maths:
Statistics 2
for Edexcel
S2.3 Continuous
distributions
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Continuous uniform distribution
Contents
Continuous uniform distribution
Approximating the binomial using a normal
Approximating the Poisson using a normal
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Continuous uniform distribution
A random variable X is said to have a continuous uniform
distribution (or rectangular distribution) over the interval
[a, b] if its probability density function has the form:
 1

f ( x)   b  a
 0
a xb
otherwise
The graph of its probability density function is as follows:
f(x)
x
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Continuous uniform distribution
Key result: If X has a continuous uniform distribution over
the interval [a, b], then
ab
E[ X ] 
2
and
1
Var[ X ] 
(b  a)2
12
Proof of E[X]: The result for E[X] follows immediately from the
symmetry of the p.d.f..
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Continuous uniform distribution
Proof of Var[X]:
b
b
1 2
(b  ab  a 2 )
3
as b3  a3  (b  a)(b2  ab  a 2 )
1
1
2
2
2
E[ X ]   x .
dx 
x
dx

ba
ba a
a
1
1
3 b
3
3



x

b

a


3(b  a)   a 3(b  a)

1 2
1
2
So, Var[ X ]  (b  ab  a )  (a  b)2
3
4
1 2
1 2
2
 (b  ab  a )  (a  2ab  b 2 )
3
4
1 2
1
2

(b  2ab  a ) 
(b  a)2
12
12
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Continuous uniform distribution
Example: A random variable Y has a continuous uniform
distribution in the interval [2, 8]. Find P(Y < μ + σ).
Using the formulae for E[X] and Var[X],
we get:
a b 28

5

2
2
2 
1
1
(8  2)2  3
(b  a)2 
12
12
  3
The required probability is P(Y < μ + σ) = P(Y < 5 + √3).
This probability is represented by the shaded area.
3 3
5

3

2

Therefore P(Y < 5 + √3) =
6
6
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Examination-style question
Examination-style question: A random variable X is given
by the probability density function f (x), where
1
5  x  15

f ( x)  10
 0 otherwise
Find:
a) E[X] and Var[X]
b) P(7 ≤ X < 10)
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Examination-style question
Solution:
X has a uniform distribution over the interval (5, 15).
ab
5  15
E
[
X
]


 10
a)
2
2
1
1
1
2
2

(
15

5
)

8
Var[ X ] 
(b  a)
12
3
12
b) The p.d.f. for X is shown on the diagram below.
The probability we require is shaded.
3
So, P(7 ≤ X < 10) =
10
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Continuous uniform distribution
Note: If X has a uniform distribution over the interval (a, b)
then the cumulative distribution function of X is:
 0
 x a
F ( x)  P( X  x)   ba

 1
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xa
a xb
xb
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Approximating the binomial using a normal
Contents
Continuous uniform distribution
Approximating the binomial using a normal
Approximating the Poisson using a normal
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Approximating the binomial using a normal
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Approximating the binomial using a normal
Calculating probabilities using the binomial distribution can
be cumbersome if the number of trials (n) is large.
Consider this example:
Introductory example:10% of people in the United
Kingdom are left-handed.
A school has 1 200 students. Find the probability that
more than 140 of them are left-handed.
Let the number of left-handed people in the school be X.
Then X ~ B[1200, 0.1].
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Approximating the binomial using a normal
The required probability is
P(X > 140) = P(X = 141) + P(X = 142) + … + P(X = 1200).
As no tables exist for this distribution, calculating this
probability by hand would be a mammoth task.
A further problem arises if you attempt to work out one of
these probabilities, for example P(X = 141):
P( X  141)  1200 C141  0.1141  0.91059
Calculators cannot calculate
the value of this coefficient –
it is too large!
One way forward is to approximate the binomial
distribution using a normal distribution.
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Approximating the binomial using a normal
Key result: If X ~ B(n, p) where n is large and p is small, then
X can be reasonably approximated using a normal distribution:
X ≈ N[np, npq]
where q = 1 – p.
There is a widely used rule of thumb that can be applied
to tell you when the approximation will be reasonable:
A binomial distribution can be approximated
reasonably well by a normal distribution
provided np > 5 and nq > 5.
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Approximating the binomial using a normal
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Approximating the binomial using a normal
A continuity correction must be applied when approximating
a discrete distribution (such as the binomial) to a continuous
distribution (such as the normal distribution).
Continuity correction:
Exact distribution: B(n, p)
P(X ≥ x)
Approximate distribution:
N[np, npq]
P(X ≥ x – 0.5)
This 0.5 is called the
continuity correction
factor.
P(X ≤ x)
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P(X ≤ x + 0.5)
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Approximating the binomial using a normal
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Approximating the binomial using a normal
Introductory example (continued): 10% of people in
the United Kingdom are left-handed.
A school has 1 200 students. Find the probability that
more than 140 of them are left-handed.
Solution:
Let the number of left-handed people in the school be X.
Then the exact distribution for X is X ~ B[1200, 0.1].
Since np = 120 > 5 and nq = 1080 > 5 we can approximate
this distribution using a normal distribution:
X ≈ N[120, 108].
np
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npq
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Approximating the binomial using a normal
So, P(X > 140) = P(X ≥ 141) → P(X ≥ 140.5)
Using continuity
correction
N[120, 108]
Standardize
N[0, 1]
140.5  120
 1.973
108
You convert 140.5 to the
standard normal distribution
using the formula:
Z
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X 

~ N[0,1].
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Approximating the binomial using a normal
Therefore P(X ≥ 140.5) = P(Z ≥ 1.973)
= 1 – Φ(1.973)
= 1 – 0.9758
= 0.0242
So the probability of there being more than 140
left-handed students at the school is 0.0242.
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Approximating the binomial using a normal
Example: It has been estimated that 15% of
schoolchildren are short-sighted. Find the probability
that in a group of 80 schoolchildren there will be
a) no more than 15 children that are short-sighted
b) exactly 10 children that are short-sighted.
Solution:
Let the number of short-sighted children in the group be X.
Then the exact distribution for X is X ~ B[80, 0.15].
Since np = 12 > 5 and nq = 68 > 5 we can approximate this
distribution using a normal distribution:
X ≈ N[12, 10.2].
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Approximating the binomial using a normal
a) So P(X ≤ 15) → P(X ≤ 15.5)
N[12, 10.2]
Using continuity correction
Standardize
N[0, 1]
15.5  12
 1.096
10.2
Therefore P(X ≤ 15.5) = P(Z ≤ 1.096)
= Φ(1.096)
= 0.8635
So the probability that no more than 15
children will be short-sighted is 0.8635.
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Approximating the binomial using a normal
b) So P(X = 10) → P(9.5 ≤ X ≤ 10.5)
Using continuity correction
Standardize
9.5  12
 0.783
10.2
N[12, 10.2]
N[0, 1]
10.5  12
 0.470
10.2
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Approximating the binomial using a normal
Therefore P(9.5 ≤ X ≤ 10.5) = P(–0.783 ≤ Z ≤ –0.470)
= P(0.470 ≤ Z ≤ 0.783)
= 0.7832 – 0.6808 = 0.1024
The probability that 10 children will be short-sighted is 0.1024.
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Examination-style question
Examination-style question:
A sweet manufacturer makes sweets in 5 colours. 25% of
the sweets it produces are red.
The company sells its sweets in tubes and in bags. There
are 10 sweets in a tube and 28 sweets in a bag. It can be
assumed that the sweets are of random colours.
a) Find the probability that there are more than 4 red
sweets in a tube.
b) Using a suitable approximation, find the probability
that a bag of sweets contains between 5 and 12 red
sweets (inclusive).
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Examination-style question
Solution:
a) Let the number of red sweets in a tube be X.
Then the exact distribution for X is X ~ B[10, 0.25].
This distribution cannot be approximated by a normal but its
probabilities are tabulated:
P(X > 4) = 1 – P(X ≤ 4)
= 1 – 0.9219
= 0.0781
So the probability that a tube contains more than 4 red
sweets is 0.0781.
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Examination style question
Solution:
b) Let the number of red sweets in a bag be Y.
Then the exact distribution for Y is Y ~ B[28, 0.25].
This distribution can be approximated by a normal since
np = 7 and nq = 21 (both greater than 5):
Y ≈ N[7, 5.25]
P(5 ≤ Y ≤ 12) → P(4.5 ≤ Y ≤ 12.5)
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npq
Using continuity correction
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Examination style question
Standardize
N[7, 5.25]
4.5  7
 1.091
5.25
12.5  7
 2.400
5.25
N[0, 1]
Therefore P(4.5 ≤ Y ≤ 12.5) = P(–1.091 ≤ Z ≤ 2.400)
= P(Z ≤ 2.400) – P(Z ≤ –1.091)
= Φ(2.400) – (1 – Φ(1.091))
= 0.9918 – (1 – 0.8623) = 0.8541
So the probability that a bag will contain between
5 and 12 red sweets is 0.8541.
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Approximating the Poisson using a normal
Contents
Continuous uniform distribution
Approximating the binomial using a normal
Approximating the Poisson using a normal
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Approximating the Poisson using a normal
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Approximating the Poisson using a normal
Key result: If X ~ Po(λ) and λ is large, then X is approximately
normally distributed:
X ≈ N[λ, λ]
Recall that the mean and variance of a Poisson distribution
are equal.
There is a widely used rule of thumb that can be applied to tell
you when the approximation will be reasonable:
A Poisson can be approximated reasonably
well by a normal distribution provided λ > 15.
Note: A continuity correction is required
because we are approximating a discrete
distribution using a continuous one.
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Approximating the Poisson using a normal
Example: An animal rescue centre finds a new home for
an average of 3.5 dogs each day.
a) What assumptions must be made for a Poisson
distribution to be an appropriate distribution?
b) Assuming that a Poisson distribution is appropriate:
i. Find the probability that at least one dog is
rehoused in a randomly chosen day.
ii. Find the probability that, in a period of 20 days,
fewer than 65 dogs are found new homes.
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Approximating the Poisson using a normal
Solution:
a) For a Poisson distribution to be appropriate we would need to
assume the following:
1. The dogs are rehoused independently of one another
and at random;
2. The dogs are rehoused one at a time;
3. The dogs are rehoused at a constant rate.
b) i) Let X represent the number of dogs rehoused on a
given day. So, X ~ Po(3.5).
P(X ≥ 1) = 1 – P(X = 0)
= 1 – 0.0302 (from tables)
= 0.9698
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Approximating the Poisson using a normal
b) ii) Let Y represent the number of dogs rehoused over a
period of 20 days. So, Y ~ Po(3.5 × 20) i.e. Po(70).
As λ is large, we can approximate this Poisson distribution
by a normal distribution:
Y ≈ N[70, 70].
P(Y < 65) → P(Y ≤ 64.5)
N[70, 70]
Standardize
N[0, 1]
64.5  70
 0.657
70
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Approximating the Poisson using a normal
P(Y ≤ 64.5) = P(Z ≤ –0.657)
= 1 – Φ(0.657)
= 1 – 0.7445
= 0.2555
So the probability that less than 65 dogs are
rehoused is 0.2555.
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Examination-style question
Examination-style question: An electrical retailer has
estimated that he sells a mean number of 5 digital radios
each week.
a) Assuming that the number of digital radios sold on
any week can be modelled by a Poisson distribution,
find the probability that the retailer sells fewer than 2
digital radios on a randomly chosen week.
b) Use a suitable approximation to decide how many
digital radios he should have in stock in order for him
to be at least 90% certain of being able to meet the
demand for radios over the next 5 weeks.
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Examination-style question
Solution:
a) Let X represent the number of digital radios sold in a week.
So X ~ Po(5).
P(X < 2) = P(X ≤ 1)
= 0.0404
(from tables).
So the probability that the retailer sells fewer than 2
digital radios in a week is 0.0404.
b) Let Y represent the number of digital radios sold in a
period of 5 weeks.
So, Y ~ Po(25).
We require y such that P(Y ≤ y) = 0.9.
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Examination-style question
Since the parameter of the Poisson distribution is large, we can
use a normal approximation:
Y ≈ N[25, 25].
P(Y ≤ y) → P(Y ≤ y + 0.5) (using a continuity correction).
N[25, 25]
Standardize
N[0, 1]
The 10% point of
a normal is 1.282
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Examination-style question
So, y  0.5  25  1.282
5
y   5  1.282   24.5
 y  30.91
So the retailer would need to keep
31 digital radios in stock.
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