Samples of 20 parts from a metal punching process are selected every hour. typically, 1% of the parts require rework. let x denote the number of parts in the sample of 20 that require rework. a process problem is suspected if x exceeds its mean by more than 3 standard deviations. (a) if the percentage of parts that require rework remains at 1%, what is the probability that x exceeds its mean by more than 3 standard deviations? (b) if the rework percentage increases to 4%, what is the probability that x exceeds 1? (c) if the rework percentage increases to 4%, what is the probability that x exceeds 1 in at least one of the next five hours of samples?

(a) 0.0169

(b) 0.189

(c) 0.651

Step-by-step explanation:

As per the question:

Total no. of samples, n = 20

Let the no. of parts requiring rework be any random variable X

The parameters for binomial distribution n = 20, p

Thus the probability distribution function in X is given by:

       (1)

where

K = 0, 1, 2, ...., 20

Now,

(a) Put the value of 'p' as 0.01 in eqn (1):

Mean is given as the product of n and p as:

E(X) =

Standard deviation is given by:

If the rework is required for 2 out of 20 parts, then:

P(X > 1.53) = 1 - P(X = 0) - P(X = 1) = 0.0169

(b) When p = 0.04, probability is given by eqn (1):

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

(c) let the no. of hours (out of 5 hrs) be represented by another random variable Y, then:

   (2)

Now, the probability is given by:

P

yess

step-by-step explanation:

idk whyy this is happening, i cant even read my notification bc i keep getting errors

~coco