Question

the quality control inspector of a production plant will reject a batch of syringes if two or more defective syringes are found in a random sample of 10 syringes taken from the batch. suppose the batch contains 5% defective syringes.

A: find the mean of a batch 1,2,3,4,5,6,7,8,9,10, where 9 and ten are the effected syringes of ten syringes

B: what is the expected number of defective syringes the inspector will find?

C: what is the probability that the batch will be accepted?

D: find the stadard deviation

Answer 1

Solution:

It is a question on Binomial distribution.
P(defective) = p = 5% = 0.05
q = 1-p = 1-0.05 = 0.95
n = 10
A) and B)
Mu or Mean or Expected number of defectives = np = 10*0.05 = 0.5

C) P(r) = nCr*q^(n-r)*p^r
Required probability = P(r </= 2) = P(r = 0)+P(r = 1) + P(r = 2)  
= 10C0*0.95^10*0.05^0 + 10C1*0.95^9*0.05^1 + 10C2*0.95^8*0.05^2
= 10C0*0.5987 * 1 * 10C1 * 0.6302 * 0.05 + 10C2 *0.6634 * 0.0025
= 0.3151 + 0.5987 + 0.0746
= 0.9885

Make the required calculations
D) Standard deviation = sqrt (npq)
= sqrt (10*0.05*0.95)
= 0.6892