Question

A manufacturer of​ push-pins sells them in packages of 400. If the overall rate of defects in the manufacturing process is known to be​ 3.1%, estimate the probability that a package of 400​ push-pins contains 15 or more defective pins.

Answer 1

Solution:

Given that,

P = 0.031

1 - P = 0.969

n = 400

Here, $X\sim$ BIN ( n , P ) that is , BIN (400 , 0.031)

then,

n*p = 12.4 > 5

n(1- P) = 387.6 > 5

According to normal approximation binomial,

X $\rightarrow$ Normal

Mean = $\mu$ = n*P = 12.4

Standard deviation = $\sigma$ =$\sqrt{}$n*p*(1-p) = $\sqrt{}$ 12.0156

We using continuity correction factor

P(X $\geq$ a ) = P(X > a - 0.5)

P(x > 14.5) = 1 - P(x < 14.5)

= 1 - P((x - $\mu$ ) / $\sigma$ < (14.5 - 12.4) / $\sqrt{}$ 12.0156)

= 1 - P(z < 0.61)

= 1 - 0.7291   

= 0.2709

Probability = 0.2709