Question

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 83 months, with a standard deviation of 99 months.

If he is correct, what is the probability that the mean of a sample of 78 computers would be less than 82.06 months? Round your answer to four decimal places.

Answer 1

Solution :

Given that,

mean = = 83

standard deviation = = 99

n = 16

= 83

  =  ( /n) = (99 / 78 ) =11.2095

P (   < 82.06 )

P ( - /) < (82.06 - 83 / 11.2095)

P ( z < - 0.94 / 11.2095 )

P ( z < -0.08 )

Using z table

= 0.4681

Probability = 0.4681