Question

In a lumber yard, there is a pile of wood that are normally distributed with mean 63.6 inches and standard deviation of 2.5 inches. A worker is randomly selecting wood until one is picked at least 72 inches. What is the probability that he first selects a piece of wood that is at least 72 inches of the third try?

Answer 1

Solution :

Given that ,

mean = = 63.6

standard deviation = = 2.5

P(x 72 ) = 1 - P(x   72)

= 1 - P[(x - ) / (72 - 63.6) / 2.5]

= 1 -  P(z 3.36)   

  Using z table,

= 1 - 0.9996

= 0.0004