Question

Suppose the life of a 1000gb solid state drive has a normal distribution with a mean of 42,000 hours and a standard deviation of 2400 hours. Suppose we have a random sample of n = 4 solid state drives, and xbar is the sample mean life of the 4 drives. What is P(41850 < xbar < 42300)

Answer 1

Given:

Mean, = 42,000 hours

Standard Deviation, = 2400 hours

n = 4

For both the limits z-score will be calculated as below

Hence, for = 41850

For = 42300

Hence,

Using the function NORMSDIST

Since, the area required is from -0.125 to 0, the value from above function is subtracted from 0.5

In excel type =0.5-NORMSDIST(-0.125)

The value is 0.0497

Since the required P(0 < Z < 0.25) in excel is calculated as the area of -infinity to 0.25, 0.5 will be subtracted from the estimated value from excel =NORMSDIST(0.25)-0.5

The value is 0.0987

Final probability is calculated is 0.0497 + 0.0987 = 0.1484

P(41850 < < 42300) = 0.1484