Question

1) It is known that in a certain manufacturing process, 1% of the products are defective. If a customer purchases 50 of these products selected at random what is probability that he receives 2 or less defective products? Epts.

Answer 1

Solution:

probability of finding defective product p = 0.01

probability of not finding a defective product q = 1 - 0.01 = 0.99

1) number of products customer purchased n = 50

Probability that he receives 2 or less defective products P(X $\leq$ 2) = P(X = 2) + P(X = 1) + P(X = 0)

Formula : P(X = k) = n!/ [k! (n-k)!] (p)^k (q)^{n - k}

  P(X $\leq$ 2) = 0.0756 +0.3056 +0.6050

= 0.9862

2) number of products customer purchased n = 200

expected value = 0* P(X =0) + 1* P(X = 1) + 2 * P(X = 2) + ...................+ 200 * P(X = 200)

= 0* (0.1340) + 1 * (0.2707) + 2 * (0.2720) + ..................+ 200 *(0)

= 1.9997

Standard deviation $\sigma$= $\sqrt{\sum x^2 . p(x) - \mu ^2}$

$\sum$x² . p(x) = 0² * (0.1340) + 1² * (0.2707) +...............+ 200² * (0)

= 5.9773

$\mu^2$ = 0.9862^{2 =} 0.9725

  $\sigma$= $\sqrt{\sum x^2 . p(x) - \mu ^2}$ = $\sqrt{5.9773 - 0.9725}$

  $\approx$1.4066