Probability Question (Statistics)?
Industrial quality control programs often include inspection of incoming materials from suppliers. If parts are purchased in large lots, a typical plan might be to select 20 parts at random from a lot and inspect them. A lot might be judged acceptable if one or fewer defective parts are found among those inspected. Otherwise, the lot is rejected and returned to the supplier. Find the probability of accepting lots that have the given percentage of defective parts. (Give the answers to three decimal places.)
(Hint: Identify success with a defective part.)
(a) 5% defective parts
P(lot will be accepted) =
(a) 15% defective parts
P(lot will be accepted) =
(a) 20% defective parts
P(lot will be accepted) =
If you could give a step-by-step id appreciate it. I’m kind of lost. It says I might have to use a table in the appendix but the only one I found that might work is the binomial probability table and I don’t know what I’m looking for on said table. (I was looking at the n=20 part on the table and assuming that the 95% at the 20 marker was the probability i entered for a, but that was wrong =( )
One Response
Julius N
30 Jun 2010
if x% of the parts are defective, you need to calculate what are the chances that you will have 0 or 1 bad parts in the selected sample.
page 831, Table 9, n=20, pi = 0.05 use the lines for x = 0 and 1
Binomial Distribution: n= 20 p= .05 µ = 1 ; σ = .9746795
– X — P(X) —— Σ P(X)
0: 0.358485917, 0.358485917,
1: 0.377353603, 0.735839519,
Binomial Distribution: n= 20 p= .15 µ = 3 ; σ = 1.596872
– X — P(X) —— Σ P(X)
0: 0.038759526, 0.038759526,
1: 0.136798332, 0.175557858,
Binomial Distribution: n= 20 p= .2 µ = 4 ; σ = 1.788854
– X — P(X) —— Σ P(X)
0: 0.011529214, 0.011529214,
1: 0.057646072, 0.069175286,
so, to put it into words, if 20% of the parts are defective then the probability of having zero or one defectives in the sample of 20 is going to be 6.9175286%