The manager of a computer retail store is concerned that his suppliers have been giving him laptop computers with lower than average quality. His research shows that replacement times for the model laptop of concern are normally distributed with a mean of 4.5 years and a standard deviation of 0.4 years. He then randomly selects records on 45 laptops sold in the past and finds that the mean replacement time is 4.4 years. Assuming that the laptop replacment times have a mean of 4.5 years and a standard deviation of 0.4 years, find the probability that 45 randomly selected laptops will have a mean replacment time of 4.4 years or less.

The probability that 45 randomly selected laptops will have a mean replacment time of 4.4 years or less is P(M<4.4)=0.0468.

Step-by-step explanation:

We have a population normally distributed with mean 4.5 years and standard deviation of 0.4 years.

Samples of size n=45 are selected from this population.

We have to calculate the probability that a sample mean is 4.4 years or less.

Then, we calculate the z-score for the sample mean M=4.4 and then calculate the probability using the standard normal distribution:

The probability that 45 randomly selected laptops will have a mean replacment time of 4.4 years or less is P(M<4.4)=0.0468.

a) justify

step-by-step explanation:

justifying is just part of everyday life. justify kind of means proof so i think the answer is a) justify.

the multiples of 15 are 15, 30,45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195,   210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375,   390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, and 735.

the factors of 15 are 1, 3, 5, and 15.

hope that , fam!

-une, your fren : )