Question

. Samples of 20 products from a production line are selected every hour. Typically, 2% of the products require improvement. Let X denote the number of products in the sample of 25 that require improvement. A production problem is suspected if X exceeds its mean by more than 3 standard deviations. (a) If the percentage of products that require improvement remains at 2%, what is the probability that X exceeds its mean by more than 3 standard deviations? (b) If the improvement percentage increases to 5%, what is the probability that X exceeds 1? (c) If the improvement percentage increases to 5%, what is the probability that X exceeds 1 in at least one of the next five hours of samples?

Answer 1

Solution:

Given that

Samples of 20 products from a production line are selected every hour. Typically, 2% of the products require improvement. Let X denote the number of products in the sample of 25 that require improvement. A production problem is suspected if X exceeds its mean by more than 3 standard deviations.

(a)

Here mean number of products that require improvement = 25 * 0.02 = 0.5

standard deviation of number of products that require improvement = sqrt [0.02 * 0.98 * 25] = 0.7

So the value that is 3 standard deviation away from the mean = 0.5 + 3 * 0.7 = 2.6

Pr(x > 2.6) = Pr(x >= 3) = BINOMDIST(x >=3 ; 25 ; 0.02) = 1- 0.9868 = 0.0132

(b)

If the improvement required percentage increases to 5%, then

Pr(X > 1) = 1 - BINOMDIST(x < =1 ; 25 ; 0.05) = 1 - 0.6424 = 0.3576

(c)

So we test 20 products for every hour

So here for a single hour

P(X > 1) = 1 - BINOMDIST(x < = 1 ; 20 ; 0.05) = 1 - 0.7358 = 0.2642

Pr(for at least one of the next five hours of samples X exceeds 1) = 1 - Pr(In none of the sample X exceeds 1)

= 1 - (1 - 0.2642)⁵ = 0.7843