Question

3. A factory worker is provided enough raw materials to construct 125 widgets in a day. Suppose each widget independently has a 16% chance of failing inspection. The worker's daily quota is 100 widgets that pass inspection. Let X be the number of widgets that pass inspection. Here is part of the distribution's table: (a) Find the expected value of X. (b) Find the third quartile. (c) Find the standard deviation 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 0.0709 0.2653 0.0831 0.3484 0.0923 0.4407 0.0969 0.5376 0.0960 0.6336 0.0895 0.7231 0.0783 0.8014 0.0641 0.8655 0.0490 0.9145 0.0347 0.9492 0.0228 0.9720 0.0138 0.9858 0.0076 0.9934 0.0038 0.9972 0.0017 0.9990 0.0007 0.9996 0.0002 0.9999 0.00011.0000 0.0000 1.0000 (d) Find the probability that exactly 100 widgets pass inspection. (e) Find the probability that the worker fails to meet her quota, i.e. P(X <100).

Answer 1

Binomial Distribution :

P(X) = nCxp*(1-P)*-*

c = pass inspection

n=125

p=1-0.16 = 0.84

nCx = n! / (x! * (n-x)!)

(a)

E = n*p

n=125, p = 0.84

E(x) = 125*0.84 = 105

E(x) = 105

b.

k = third quartile

P(x<=k) = 0.75 {third quartile condition}

P(x<=k) = P(1) + P(2) + .... + P(k) = 0.75

P(x<=107) = 0.7231

P(x<=108) = 0.8014

therefore third quartile = 108

c.

in binomial distribution

var = n*p*(1-p)

var(x) = 125*0.84*(1-0.84) = 16.8

SD = var^0.5 = 4.0987

SD = 4.0987

d.

x = 100

P(100) = (125C100)*(0.84^100)*(0.16^25) = 0.04416

P(100) = 0.04416

e.

P(x < q) = P(1) + P(2) + .... + P(q-1)

P(x<100) = P(1) + P(2) + .... + P(99)

P(x<100) = 0.092827

P.S. (please upvote if you find the answer satisfactory)