Statistics 2

Statistics 2

Factorial Notation

• Factorials
• n objects in a row can be arranged in n! ways.
• n! = n(n – 1)(n – 2)(n – 3) × .... × 4 × 3 × 2 ×1
• Note that 0! is defined to be 1. This fits in with formulae for combinations.
• Combinations
• The number of ways we can choose r objects from a total of n objects, where the order does not
• matter, is called the number of combinations of r objects from n and is written as
• nCr = n(n-1)(n-2)...(n-r+1)/r! or nCr = n!/(n-r)!r!
• We can think of this as n choose r.
• Example: Find the number of hands of 4 cards which can be dealt from a pack of 10.
• Solution: In a hand of cards the order does not matter, so this is just the number of combinations of 4
• from 10, or 10 choose 4
• 10C4 = 10x9x8x7/4! = 210

Binomial Thereom

• To write down the expansion of (p + q)
• 6
• We write down the terms in a logical order then use the numbers in the ‘6th’ row of the triangle.
• Binomial coefficients and combinations
• The binomial coefficients ቀ
• ݊
• ݎቁ are equal to the number of combinations nCr
• ⇒ = nCr = n!/(n-r)!r!

Binomial Distribution (n,p)

• Conditions for a Binomial Distribution
• - A single trial has exactly two possible outcomes – success and failure.
• - This trial is repeated a fixed number, n, times.
• - The n trials are independent.
• T
• - The probability of success, p, remains the same for each trial.
• The probability of success in a single trial is usually taken as p and the probability of failure as q.
• Note that p + q = 1.

Cumulative binomial probability tables

• Example: For X ~ B(30, 0⋅35), find the probability that 7
• Solution: A moment’s thought shows that we need
• P(X = 8, 9, 10, 11 or 12)
• = P(X ≤ 12) – P(X ≤ 7) = 0⋅7802 – 0⋅1238, using tables for n = 30, p = 0⋅35
• = 0⋅6564 to 4 D.P. from tables
• Example:
• A bag contains a large number of red and white discs, of which 85% are red. 20 discs are
• taken from the bag; find the probability that the number of red discs lies between 12 and 17
• inclusive.
• Solution: As there is a large number of discs in the bag, we can assume that the probability of a red
• disc remains the same for each trial, p = 0⋅85.
• Let X be the number of red discs ⇒ X ~ B(20, 0⋅85)
• We now want P(12 ≤ X ≤ 17).
• At first glance this looks simple until we realise that the tables stop at probabilities of 0⋅5.
• We need to consider the number of white discs, Y ~ B(20, 0⋅15), where 0⋅15 = 1 – 0⋅85,
• For 12 ≤ X ≤ 17
• we have X = 12, 13, 14, 15, 16 or 17
• Y = 8, 7, 6, 5, 4 or 3
• ⇒ P(12 ≤ X

Mean and Variance

• If X ~ B(n, p)
• then
• the expected mean is E[X] = μ = np,
• the expected variance is Var[X] = σ
• 2
• = npq = np(1 – p).
• Example: A coin is spun 100 times.
• Find the expected mean and variance of the number of heads.
• Solution: X ∼ B(100, ଵ
• ଶ
• )
• ⇒ μ = np = 100 × 1/2 = 50
• and σ
• 2
• = npq = 100 × 1/2 × 3/4 = 37⋅5
• ⇒ σ = √37 · 5 = 6⋅1237…
• μ – 2σ = 50 – 12⋅24… = 38,
• μ + 2σ = 50 + 12⋅24… = 62
• So we would expect that a probability of about 0⋅95 that the number of heads lies between 38 and 62,
• inclusive. This assumes that the probability distribution is approximately Normal.
• Example: It is believed that 35% of people like fish and chips. A survey is conducted to verify this.
• Find the minimum number of people who should be surveyed if the expected number of people
• who like fish and chips is to exceed 60.
• Solution: If X is the number of people who like fish and chips in a sample of size n,
• then X ∼ B(n, 0⋅35).
• E[x] = μ = np = 0.35n > 60
• ⇒ n > 60/0.35 = 171.4285714
• ⇒ n = 172, since the expected mean has to exceed 60.

Poisson Distribution

• Conditions for a Poisson distribution
• Events must occur
• Singly – two cannot occur simultaneously
• Uniformly – at a constant rate
• Independently and randomly – the occurrence of one event does not influence the occurrence of
• another

Folie 8

• As before, the term probability distribution means
• the set of all possible outcomes (in this case the values of x = 0, 1, 2, …, n)
• together with their probabilities.
• Notice that in a Poisson distribution, x can take any positive or zero integral value, no matter how
• large.
• In practice, the probabilities of the ‘larger’ values will be very, very small.
• Finding probabilities for the Poisson distribution is very similar to finding probabilities for the
• Binomial
• Example: Cars pass a particular point at a rate of 5 cars per minute.
• (a) Find the probability that exactly 4 cars pass the point in a minute.
• (b) Find the probability that between at least 3 but fewer than 8 cars pass in a particular
• minute.
• (c) Find the probability that more than 8 cars pass in 2 minutes.
• (d) Find the probability that more than 3 cars pass in each of two separate minutes.
• Solution: Let X be the number of cars passing in a minute, then X ∼ PO(5)
• (a) X ∼ PO(5)
• ⇒ P(X = 4) = ହరൈ ௘షఱ
• ସ!
• = 0⋅175467369768 = 0⋅175 to 3 S.F.
• or P(X = 4) = P(X ≤ 4) – P(X ≤ 3) = 0⋅4405 – 0⋅2650 = 0⋅1755 to 4 D.P.
• (b) P(at least 3 but fewer than 8) = P(3 ≤ X (c) We know that the Poisson distribution is uniform, so if a mean of 5 cars pass each minute,
• it means that a mean of 10 cars pass in a two minute period.
• Thus, if Y is the number of cars passing in two minutes
• Y ∼ PO(10), and we need
• P(Y > 8) = 1 – P(Y ≤ 8) = 1 – 0⋅3328 = 0⋅6672 to 4 D.P.
• (d) For probability of more than 3 in one 1 minute period, we have X ∼ PO(5)
• ⇒ P(X > 3) = 1 – P(X ≤ 2) = 1 – 0⋅1247 = 0⋅8753 from tables
• ⇒ probability of this happening in two separate minutes
• is 0⋅8753^2
• = 0⋅76615009 = 0⋅766 to 3 S.F.

Mean and Variance

• If X ~ PO(λ) then it can be shown that
• the expected mean is E[X] = μ = λ
• the expected variance is Var[X] = σ
• 2
• = λ.
• This means that if the set of n trials were to be repeated a large number, n, times and the number of
• occurrences recorded each time, x1, x2, x3, x4 ..... xn
• Example: In producing rolls of cloth there are on average 4 flaws in every 10 metres of cloth.
• (a) Find the mean number of flaws in a 30 metre length.
• (b) Find the probability of fewer than 3 flaws in a 6 metre length.
• (c) Find the variance of the number of flaws in a 15 metre length.
• Solution: Assuming a Poisson distribution – flaws in the cloth occur singly, independently,
• uniformly and randomly.
• (a) If the mean number of flaws in 10 metres is 4,
• then the mean number of flaws in 30 metre lengths is 3 × 4 = 12.
• (b) If there are 4 flaws on average in a 10 metre length there will be 6/10 × 4 = 2⋅4 flaws on
• average in a 6 metre length.
• If X is the number of flaws in a 6 metre length then X ∼ PO(2.4).
• P(X (c) If the mean number of flaws in 10 metre lengths is 4, then the mean number of flaws in 15
• metre lengths will be
• λ = 15/10 × 4 = 6
• Since, in a Poison distribution, the variance equals the mean
• the variance is 6.

Poisson Approximation

• Binomial B(n, p) for small p
• If in the Binomial distribution B(n, p) p is ‘small’ and n is ‘large’, then we can approximate by a
• Poisson distribution with mean λ = np, PO(np).
• Notice that when p is ‘small’ and n is ‘large’,
• the expected variance of B(n, p) is npq ≈ np since q = 1 – p ≈ 1
• and so expected mean ≈ the variance.
• In a Poisson distribution, the expected mean = the variance, so the approximation is suitable.
• The approximation is also suitable if q is ‘small’ and n is ‘large’.