© boardworks ltd 20061 of 39 © boardworks ltd 2006 1 of 39 a2-level maths: statistics 2 for...

© Boardworks Ltd 20061 of 39 © Boardworks Ltd 20061 of 39

A2-Level Maths: Statistics 2for Edexcel

S2.3 Continuous distributions

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Continuous uniform distribution


Approximating the binomial using a normal

Approximating the Poisson using a normal


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

0 otherwise

a x bf x b a

f(x)

The graph of its probability density function is as follows:

x



[ ]E2

a bX

and

Proof of E[X]: The result for E[X] follows immediately from the symmetry of the p.d.f..

[ ] ( )21Var

12X b a


Key result: If X has a continuous uniform distribution over the interval [a, b], then



( ) ( )

3 3 31 1

3 3

b

ax b a

b a b a

( ) ( )2 2 2 21 12

3 4b ab a a ab b

[ ] .2 2 21 1b b

a a

E X x dx x dxb a b a

Proof of Var[X]:

( )2 21

3b ab a

So, [ ] ( ) ( )2 2 21 1Var

3 4X b ab a a b

( )2 212

12b ab a

( )( )3 3 2 2 b a b a b ab a as

( )21

12b a


Example: A random variable Y has a continuous uniform distribution in the interval [2, 8]. Find P(Y < μ + σ).


2

a b

Using the formulae for E[X] and Var[X],we get:

( )2 21

12b a

The required probability is P(Y < μ + σ) = P(Y < 5 + √3).This probability is represented by the shaded area.

5 3 2

6

2 85

2

3

( )218 2 3

12

Therefore P(Y < 5 + √3) =3 3

6


Examination-style question: A random variable X is given by the probability density function f (x), where

( )1

5 15100 otherwise

xf x

Find:

a) E[X] and Var[X]

b) P(7 ≤ X < 10)

Examination-style question


Solution:X has a uniform distribution over the interval (5, 15).

a)


[ ]E2

a bX

[ ] ( )21Var

12X b a

b) The p.d.f. for X is shown on the diagram below. The probability we require is shaded.

So, P(7 ≤ X < 10) =3

10

5 1510

2

( )21 115 5 8

12 3


Note: If X has a uniform distribution over the interval (a, b) then the cumulative distribution function of X is:

( ) ( )

0

P

1

x ab a

x a

F x X x a x b

x b



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Calculating probabilities using the binomial distribution can be cumbersome if the number of trials (n) is large.

Consider this example:


Let the number of left-handed people in the school be X.

Then X ~ B[1200, 0.1].

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.


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):

( ) . .1200 141 1059141P 141 C 0 1 0 9X

One way forward is to approximate the binomial distribution using a normal distribution.


Calculators cannot calculatethe value of this coefficient –

it is too large!


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.


Continuity correction:

Exact distribution: B(n, p)

P(X ≥ x) P(X ≥ x – 0.5)

P(X ≤ x) P(X ≤ x + 0.5)


Approximate distribution: N[np, npq]

This 0.5 is called thecontinuity correction

factor.

A continuity correction must be applied when approximating a discrete distribution (such as the binomial) to a continuous distribution (such as the normal distribution).


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 npq


So, P(X > 140) = P(X ≥ 141) → P(X ≥ 140.5)

..

140 5 1201 973

108


StandardizeN[120, 108]

Using continuity correction

You convert 140.5 to the standard normal distribution

using the formula:

~ [ ].N 0,1X

Z

N[0, 1]



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.


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


Solution:

Let the number of short-sighted children in the group be X.


Since np = 12 > 5 and nq = 68 > 5 we can approximate this distribution using a normal distribution:

X ≈ N[12, 10.2].

a) no more than 15 children that are short-sighted

b) exactly 10 children that are short-sighted.


a) So P(X ≤ 15) → P(X ≤ 15.5)

..

.

15 5 121 096

10 2

Therefore P(X ≤ 15.5) = P(Z ≤ 1.096)

= Φ(1.096)

= 0.8635



StandardizeN[12, 10.2] N[0, 1]

So the probability that no more than 15 children will be short-sighted is 0.8635.


b) So P(X = 10) → P(9.5 ≤ X ≤ 10.5)

..

.

9 5 120 783

10 2


..

.

10 5 120 470

10 2


N[12, 10.2]

Standardize

N[0, 1]



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.


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).


Solution:

a) Let the number of red sweets in a tube be X.


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.



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)

Examination style question

npq



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)

So the probability that a bag will contain between 5 and 12 red sweets is 0.8541.

..

.

12 5 72 400

5 25

..

.

4 5 71 091

5 25

Examination style question

N[7, 5.25] N[0, 1]

= 0.8541

Standardize


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Key result: If X ~ Po(λ) and λ is large, then X is approximately normally distributed:

X ≈ N[λ, λ]

There is a widely used rule of thumb that can be applied to tell you when the approximation will be reasonable:

Recall that the mean and variance of a Poisson distribution are equal.

Note: A continuity correction is required because we are approximating a discrete

distribution using a continuous one.


A Poisson can be approximated reasonably well by a normal distribution provided λ > 15.


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.


Solution:a) For a Poisson distribution to be appropriate we would need to

assume the following:

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


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) 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)

..

64 5 700 657

70


StandardizeN[70, 70] N[0, 1]



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.


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.


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).


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.

So the probability that the retailer sells fewer than 2 digital radios in a week is 0.0404.


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] N[0, 1]Standardize

The 10% point of a normal is 1.282



..

0 5 251 282

5

y So,

.30 91y

So the retailer would need to keep 31 digital radios in stock.

. .5 1 282 24 5y

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