Question

Assuming that the laptop replacement times have a mean of 3.3 years and a standard deviation of 0.6 years, find the probability that 41 randomly selected laptops will have a mean replacement time of 3.1 years or less.

Answer 1

Answer:

Required Probability = 0.01659

Step-by-step explanation:

We are given that the laptop replacement times have a mean of 3.3 years and a standard deviation of 0.6 years i.e.; $\mu =$ 3.3 years   and  $\sigma$ = 0.6 years

Sample of 41 is taken i.e. n = 41.

The Z score probability is given by;

      Z = $\frac{Xbar -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

Let X bar = mean replacement time

So, probability that randomly selected laptop have a mean replacement time of 3.1 years or less = P(X bar <= 3.1 years)

P(X bar <= 3.1) = P( $\frac{Xbar -\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{3.1 -3.3}{\frac{0.6}{\sqrt{41} } }$ ) = P(Z <= -2.13) = 1 - P(Z < 2.13)

                                                             = 1 - 0.98341 = 0.01659

Therefore, probability that 41 randomly selected laptops will have a mean replacement time of 3.1 years or less is 0.01659 .