Question

Statisticians use sampling plans to either accept or reject batches or lots of material. Suppose one of these sampling plans involves sampling independently 10 items from a lot of 100 items in which 12 are defective. Let X be the random variable defined as the number of items found defective in the sample of 10. In this case, the random variable takes on the values 0, 1, 2,..., 9, 10.

Answer 1

Answer:

x= {0,1,2,3,4,5,6,7,8,9,10}

Step-by-step explanation:

As the random variable takes the values from 0 ----10 it can be written as

x= {0,1,2,3,4,5,6,7,8,9,10}

The probability of the defective and non defective items would be

P ( defective) = 12/ 100 = 0.12

P ( non defective) = 1-0.12= 0.88

Since the events are independent the probability of defective items in the randomly 10 selected items would be given

P (1/10) * 0.12 +P ( 2/10) * 0.12 + P(3/10)*0.12 ,---------,+P (10/10) * 0.12

Similarly the probability of non defective item s would be

P (1/10) * 0.88 +P ( 2/10) * 0.88 + P(3/10)*0.88 ---------P (10/10) * 0.88

That is the individual probability is multiplied with the probability of the defective or non defective items to get the total probability of defective and non defective items as the events are independent.