Question

calculate

PART A: The manager of a computer retails 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 3.7 years and a standard deviation of 0.5 years. He then randomly selects records on 44 laptops sold in the past and finds that the mean replacement time is 3.6 years.

Assuming that the laptop replacement times have a mean of 3.7 years and a standard deviation of 0.5 years, find the probability that 44 randomly selected laptops will have a mean replacement time of 3.6 years or less.
P(M < 3.6 years) =  
Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 256.8-cm and a standard deviation of 0.8-cm. For shipment, 15 steel rods are bundled together.

Find the probability that the average length of a randomly selected bundle of steel rods is between 256.2-cm and 257-cm.
P(256.2-cm < M < 257-cm) =

Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

PART C: A manufacturer knows that their items have a normally distributed lifespan, with a mean of 6.4 years, and standard deviation of 0.5 years.

If 24 items are picked at random, 8% of the time their mean life will be less than how many years?

Give your answer to one decimal place.

PART B:

Answer 1

Solution:-

Part A) The probability that 44 randomly selected laptops will have a mean replacement time of 3.6 years or less is 0.0923.

Mean = 3.7, S.D = 0.50, n = 44

x = 3.6

By applying normal distribution:-

z = -1.327

P(z < - 1.327) = 0.0923

The probability that the average length of a randomly selected bundle of steel rods is between 256.2-cm and 257-cm is 0.8317

Mean = 256.8, S.D = 0.80, n = 15

x₁ = 256.2

x₂ = 257

By applying normal distribution:-

z₁ = - 2.905

z₂ = 0.968

P( -2.905 < z < 0.968) = P(z > -2.905) - P(z > 0.968)

P( -2.905 < z < 0.968) = 0.9982 - 0.1665

P( -2.905 < z < 0.968) = 0.8317

The probability that the average length of a randomly selected bundle of steel rods is between 256.2-cm and 257-cm is 0.8317