Question

Whe to apply the central limit theorem to make various estimates. Required: a. Compute the standard error of the sampling distribution of sample meansi (Round your answer to 2 decimal places.) b. What is the chance HLI will find a sample mean between 4.7 and 5.9 hours? (Round your z and standard error values to 2 decimal places. Round your intermediate and final answer to 4 decimal places.) c. Calculate the probability that the sample mean will be between 5.1 and 5.5 hours. (Round your z and standard errot values to 2 decimal places. Round your intermediate and final answer to 4 decimal places.) C. Cuiculate the probability that the stample mean will be between 5.1 and 5.5 hours. (Aound your z and standard error values ta 2 decimal places. Round your Intermediate and final answer to 4 decimal places.) d. How strange would it be to obtain a sample mean greater than 7.60 hours? This is very unlikely. This is very likely.

Answer 1

a. To find the standard error of the sampling distribution of sample means:

Standard deviation = sqrt(Variance of the population)

Since the population standard deviation is not given, we assume it is 1.

Standard error = (Standard deviation) / sqrt(n)

= (1) / sqrt(100)

= 0.01 (rounded to 2 decimal places)

b.

Standard error = 0.01 (from part a)

z = (4.7 - mean) / 0.01

= (4.7 - 5) / 0.01

= -0.3 (rounded to 2 decimal places)

Chance that sample mean is between 4.7 and 5.9 hours

= P(z > -0.3) + P(z < 0.3)

= 0.762 + 0.761

= 0.7524 (rounded to 4 decimal places)

c.

Standard error = 0.01 (from part a)

z = (5.1 - mean) / 0.01

= 0.1 (rounded to 2 decimal places)

Chance that sample mean is between 5.1 and 5.5 hours

= P(z > 0.1) + P(z < -0.1)

= 0.4583 + 0.4603

= 0.4593 (rounded to 4 decimal places)

d.

Standard error = 0.01 (from part a)

z = (7.60 - mean) / 0.01

= 3 (rounded to 2 decimal places)

Chance that sample mean is greater than 7.60 hours

= P(z > 3)

= 0 (rounded to 4 decimal places)

This would be very unlikely.