Question #286169



6. In the manufacture of car tyres, a particular production process is know to yield 10 tyres 

with defective walls in every batch of 100 tyres produced. From a production batch of 100 tyres, a sample of 4 is selected for testing to destruction. Find: 


i) the probability that the sample contains 1 defective tyre. (ii) the expectation of the number of defectives in samples of size 4. (iii) the variance of the number of defectives in samples of size 4.


1
Expert's answer
2022-01-10T15:57:48-0500

Let X= the number of defective tyres.

The hypergeometric distribution of the random variable X:

h(x;n,M,N)=(Mx)(NMnx)(Nn)N=100M=10n=4h(x;n,M,N)= \frac{\binom{M}{x} \binom{N-M}{n-x}}{\binom{N}{n}} \\ N=100 \\ M= 10 \\ n=4

(i)

P(X=1)=(101)(1001041)(1004)=0.30P(X=1) = \frac{\binom{10}{1} \binom{100-10}{4-1}}{\binom{100}{4}} = 0.30

(ii)

E(X)=x=04xP(X=x)=n×MN=4×10100=0.4E(X) = \sum^4_{x=0} x P(X=x) \\ = \frac{n \times M}{N} \\ = \frac{4 \times 10}{100} = 0.4

Hypergeometric distribution mean formula.

(iii)

Var(X)=n×MN×(NM)N×(Nn)N1=4×10100×90100×9699=0.35Var(X) = n \times \frac{M}{N} \times \frac{(N-M)}{N} \times \frac{(N-n)}{N-1} \\ = 4 \times \frac{10}{100} \times \frac{90}{100} \times \frac{96}{99} = 0.35


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