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.

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.; 3.3 years   and   = 0.6 years

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

The Z score probability is given by;

      Z = ~ 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( <= ) = 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 .

the answer is 844

step-by-step explanation:

it is 844 because if you add 836.50 + 7.5 you will get 844

the answer is b.

step-by-step explanation:

identity property is where 0+whatever number=whatever number. basically 0+ a number.