Strategic Marketing

Q.4

The mean inside diameter of a sample of 200 washers produced by a machine is 5.02mm and the Std. deviation is 0.05mm. The purpose for which these washers are intended allows a maximum tolerance in the diameter of 4.96mm to 5.08mm, otherwise the washers are considered defective. Determine the percentage of defective the washers produced by the machine, assuming the diameters are normally distributed.

If X is normal random variable with mean μ and standard deviation σ ,then

Z=X-μσ

is a standard normal random variable and hence

P(x1<X<x2) = P(x1-μσ<Z<x2-μσ)

Answer.

1-4.96<X<5.08

= 1-P( 4.96-5.020.05< X-μσ<5.08-5.020.05)

= 23.02%

Q.10

A pleasant hunter brings down 75% of the birds he shoots at. What is the probability that at least 3 of the next 5 pheasants shot at will escape? If X represents the number of pheasants that escape when 5 pheasants are shot at, find the probability distribution of X.

Answer.

In a binomial experiment with a constant probability p of success at each trial, the probability distribution of the binomial random variable X, the number of successes in n independent trials, is called the binomial distribution.

Notation: X ~ b(n, p)

X ~ b (5, 14)

Probability

= 1-P(X=0)-P(X=1)-P(X=2)

= 1-C05(o.25)0(0.75)5-C45(o.25)1(0.75)4-C25(o.25)2(0.75)3

=0.1035