Continuous uniform distribution PPT - fhsmaths

A2-Level Maths: Statistics 2 for Edexcel

S2.3 Continuous distributions This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 39

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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 o th e rw ise

The graph of its probability density function is as follows: f(x)

x 3 of 39

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

ab 2

and V a r[ X ] 

1

(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

E[ X ]  2



1

2

x .

a

ba

1

dx 

b

1 ba



2

x dx

a

1 3 3    x  b  a     a 3 (b  a ) 3 (b  a )



1

3

b

( b  ab  a ) 2

2

as

b  a  ( b  a )( b  ab  a ) 3

3

2

2

3

So, V a r [ X ]   

1

( b  ab  a )  2

3 1

1

(a  b)

2

4 ( b  ab  a ) 

3 1 12

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2

2

2

1

( a  2 ab  b ) 2

2

4 ( b  2 ab  a )  2

2

1

(b  a )

2

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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:  



2



ab



28

2

1 12

5

2

(b  a )

2



1

(8  2 )  3 2

12

  

3

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

5

3 2 6

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

3

3

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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  f ( x)  10  0 

5  x  15 o th e rw ise

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). E[ X ] 

a)

ab 2

V a r[ X ] 



5  15

 10

2

1 12

(b  a )  2

1 12

(1 5  5 )  8 2

1 3

b) The p.d.f. for X is shown on the diagram below. The probability we require is shaded. So, P(7 ≤ X < 10) =

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3 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

C 141  0 . 1

141

 0 .9

1059

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. 13 of 39

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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) 16 of 39

P(X ≤ x + 0.5) © Boardworks Ltd 2006

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 18 of 39

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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

1 4 0 .5  1 2 0

N[0, 1]

 1 .9 7 3

108

You convert 140.5 to the standard normal distribution using the formula: Z  19 of 39

X 



~ N[ 0,1]. © Boardworks Ltd 2006

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]. 21 of 39

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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] 1 5 .5  1 2

 1 .0 9 6

1 0 .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. 22 of 39

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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  1 2

  0 .7 8 3

1 0 .2

N[12, 10.2]

N[0, 1] 1 0 .5  1 2

  0 .4 7 0

1 0 .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 .0 9 1

N[0, 1]

5 .2 5 1 2 .5  7

 2 .4 0 0

5 .2 5

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. 28 of 39

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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. 31 of 39

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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 33 of 39

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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

6 4 .5  7 0

N[0, 1]

  0 .6 5 7

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. 37 of 39

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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 38 of 39

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Examination-style question So, y  0 . 5  2 5  1 . 2 8 2 5 y   5  1 .2 8 2   2 4 .5

 y  3 0 .9 1

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

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