statistics and probability

Question

The demand of replacement of windscreens at a certain garage occurs randomly at an average rate of 10 per week. 11 Determine the probability that not less than 3 windscreens are demanded in a week. (2 marks) 12 If the garage is open Monday to Saturday, determine the probability that the garage gets any demand for windscreen on Monday. (3 marks) 13 The windscreen manufacturer uses glass which contains random flaws at an average of 5 per 10m2 . A windscreen of area 0.5m2 is chosen at random, determine the probability that the windscreen has at most 1 flaw. (4 marks) 14 A random sample of 5 such windscreens was taken. Find the probability that 2 of them had at most 1 flaw. (HINT: use your answer from 13) (

Answer 1

These problems use the Poisson distribution (with the last also using the binomial)

With the Poisson, with rate λ, P(x) = e^-λ * λ^x/x!.

Then, P(X <= x) = P(0) + P(1) + ... P(x), and P(X >= x) = 1 - P(X < x) = 1 - P(X <= x - 1)

Note that packages provide both P(X = x) - the probability mass function, and P(X <= x), the cumulative probability.

Thus, I will show two ways to get the answer to the first problem.

11 λ = 10 P(not less than 3 windscreens needed) = 1 - P(less than 3 windscreens needed = 1 - P(2 or less windscreens needed)

This equals 1 - (P(0) + P(1) + P(2)) = 1 - (e^-λ * λ⁰/0! + e^-λ * λ¹/1! + e^-λ * λ²/2!) =

1 - (e^-10 * 1 + e^-10 * 10 + e^-10 * 100/2) = 1 - e^-10 * (1 + 10 + 50) = 1 - 61 e^-10 =

1 - .0028 =

.9972

Alternatively, from Excel, 1 - poisson.dist(2,10,1) = .9972. The 1 in parentheses says that you are looking for the cumulative probability

12 As the garage is open Monday-Saturday equals 6 days/week, and 10 is the weekly rate, if we assume equal rates for each day, the rate for Monday is 10/6 = 5/3

P(any demand on Monday) = P(X >= 1) = 1 - P(X = 0) = 1 - e^-λ * λ⁰/0! = 1 - e^-5/3 * 5/3⁰/0! =

1 - e^-5/3 * 1 = 1 - e^-5/3 = .8111

We can also use 1 - poisson.dist(0,5/3,0) = .8111 (With 0, it does not matter if you have 0 or 1 in the last place, as P(X <= 0) = P(X = 0)

13 Flaw rate is 5/10m² (Did you type this correctly? Is this 5/10 or 5/1.0?)

Then, in .5m² , the flaw rate is .5m² * 5/10m² = .25

Then, the probability of at most one flaw is

P(X <= 1) = P(0) + P(1) = e^-λ * λ⁰/0! + e^-λ * λ¹/1! = e^-.25 * .25⁰/0! + e^-.25 * .25¹/1! =

e^-.25 (1 + .25) = 1.25e^-.25 = .9735

Alternatively, poisson.dist(1,.25,1) = .9735

14 P(X = 2 from a sample of 5, each with probability 1.25 e^-.25 , is, from the binomial distribution formula,

P(X = 2 | n = 5, p = 1.25 e^-.25 ) =

C(5,2)p²(1-p)^5-2 = from the previous problem

10*.9735²(1-.9735)³ = .000176 To four decimal places, this is .0002

We can also use Excel's binomial distribution formula

=binom.dist(2,5,.9735,0) = .000176 = .0002