Question

Houseflies have short lifespans. Males of a certain species have lifespans that are strongly skewed to the right with a mean of 26 days and a standard deviation of 12 days. a) Explain why you cannot determine the probability that a given male housefly will live less than 24 days. b) Can you estimate the probability that the mean lifespan for a sample of 5 male houseflies is less than 24 days? Explain. c) A biologist collects a random sample of 65 of these male houseflies and observes them to calculate the sample mean lifespan. Describe the sampling distribution of the mean lifespan for samples of size 65. d) What is the probability that the mean lifespan for the sample of 65 houseflies is less than 24 days? e) What is the mean lifespan for the top 15% of samples of size 65?

Answer 1

Solution:

Houseflies have short lifespans. Males of a certain species have lifespans that are strongly skewed to the right with a mean of 26 days and a standard deviation of 12 days.

We are given

Mean = µ = 26

SD = σ = 12

a) Explain why you cannot determine the probability that a given male housefly will live less than 24 days.

We cannot determine the probability that a given male housefly will live less than 24 days because lifespan are not normally distributed and given data for lifespan is right skewed in nature. We only use z-score approximation process when variable follows a normal or approximately normal distribution.

b) Can you estimate the probability that the mean lifespan for a sample of 5 male houseflies is less than 24 days? Explain.

Yes, we can estimate the probability that the mean lifespan for a sample of 5 male houseflies is less than 24 days because sampling distribution of sample mean follows a normal distribution although the data for lifespan do not follows a normal distribution. This is the property of sampling distribution. Sampling distribution of any sample statistic follows an approximate normal distribution although the variable follows or not a normal distribution.

c) A biologist collects a random sample of 65 of these male houseflies and observes them to calculate the sample mean lifespan. Describe the sampling distribution of the mean lifespan for samples of size 65.

Estimate for Mean of sampling distribution = µ = 26

Estimate for standard deviation of sampling distribution = σ/sqrt(n)

We are given σ = 12, n = 65

Estimate for standard deviation of sampling distribution = 12/sqrt(65)

Estimate for standard deviation of sampling distribution = 1.488416815

d) What is the probability that the mean lifespan for the sample of 65 houseflies is less than 24 days?

We have to find P(Xbar<24)

Z = (Xbar - µ) / [σ/sqrt(n)]

Z = (24 – 26) / [12/sqrt(65)]

Z = -2/ 1.488416815

Z = -1.343709625

P(Z< -1.343709625) = 0.089521152 (by using z-table/excel/Ti-83/84 calculator)

P(Xbar<24) = 0.089521152

Required probability = 0.089521152

e) What is the mean lifespan for the top 15% of samples of size 65?

Solution:

Formula for required lifespan X is given as below:

X = µ + Z*[σ/sqrt(n)]

Z for top 15% is given as below:

Z = 1.036433389

(by using z-table/excel/Ti-83/84 calculator)

X = 26 + 1.036433389*[12/sqrt(65)]

X = 26 + 1.036433389*1.488416815

X = 27.54264488

Required mean lifespan = 27.54 days