Urgent help please with this question, thank you.

Question

1. Suppose that 5% of the first batch of engines off a new production line have flaws. An inspector randomly selects six engines for testing.

a) Show the probability distribution for the number of flawed engines in the sample.

It says to use probability distribution, but the answer shows to use binomial distribution? I'm so confused, how do you know when to use it in any problem?



b) What is the expected number of flawed engines in the sample?

Please explain in detail what you have to do for this question, thank you.

Answer 1

P(x = 0) = C(6,0)p⁰(1 - p)⁶ = 1·0.05⁰·0.95⁶ = .7350918906

P(x = 1) = C(6,1)p¹(1- p)⁵ = 6·0.05·0.95⁵ = .2321342813

P(x = 2) = C(6,2)p²(1- p)⁴ = 15·0.05²·0.95⁴ = .0305439844

P(x = 3) = C(6,3)p³(1 - p)³ = 20·0.05³·0.95³ = .0021434375

P(x = 4) = C(6,4)p⁴(1 - p)² = 15·0.05⁴·0.95² = .000084609375

P(x = 5) = C(6,5)p⁵(1 - p)¹ = 6·0.05⁵·0.95 = .00000178125

P(x = 6) = C(6,6)p⁶(1 - p)⁰ = 1·0.05⁶·0.95⁰ = .000000015625

Expected value M = ∑xP(x) = 1·0.2321342813 + 2·0.0305439844 + 3·0.0021434375 + 4·0.000084609375 +

5·0.00000178125 + 6·0.0000000015625 = .3

Answer 2

It's binomially distributed (as there are two possible outcomes for an engine, flawed or not flawed)

n = 6 as 6 engines are tested

p = 5% = 0.05 = ptoorobability of a flaw

So distribution is B (n,p) = B (6, 0.05)

[ binomial distribution is one example of a probability distribution. Others are normal. poisson etc ]

(b) Expected number in a binomial distribution = np = 0.05 x 6 = 0.30

What this means is if you take a large number of 6 engines samples, the average number of flawed engines in a sample will be 0.3. [ So if you took 1000 6 engine samples, and added up the number of flawed engines in the 1000 samples you would expect there to be 0.3 x 1000 = 300 flawed engines out of the 6000 tested ]