Question

The amounts of electricity bills for all households in a particular city have approximately normal distribution with a mean of $140 and a standard deviation of $30. A researcher took random samples of 25 electricity bills for a certain study.

a. Based on this information, what is the expected value of the mean of the sampling distribution of mean?

b. What is the standard error of the sampling distribution of mean?

c. If his one sample of 25 bills has an average cost of $155, what is the z-score that represents the location of this value in the sampling distribution?

d. What is the probability of getting this result or greater (enter as a percentage - if your percentage is less than one full percent enter a zero before the decimal point)?

e. Assuming that z-scores greater than +2 or less than -2 are unlikely to occur due to sampling error alone, is this a likely or unlikely result?

Answer 1

By Central limit theorem If X follows a normal distribution with mean and standard deviation then  sampling distribution of sample mean (sample size n)   with mean =   and standard error =  

X:  amounts of electricity bills for household in a particular city

X follows normal distribution with mean $140 and standard deviation 30

Number of electricity bill in a random sample : n= 25

By Central limit theorem, sampling distribution of sample mean with mean(expected value )  = 140 and standard error : = ==6

a. Based on this information, what is the expected value of the mean of the sampling distribution of mean

expected value of the mean of the sampling distribution of mean = 140

b. What is the standard error of the sampling distribution of mean?

standard error of the sampling distribution of mean = = ==6

standard error of the sampling distribution of mean = 6

c. If his one sample of 25 bills has an average cost of $155, what is the z-score that represents the location of this value in the sampling distribution

z-score = (155-mean)/standard error=(155-140)/6 = 15/6 = 2.5

z-score that represents the location of this value (155) in the sampling distribution = 2.5

d. What is the probability of getting this result or greater

probability of getting this result or greater = P(>155) = P(Z>2.5)

P(Z>2.5)=1-P(Z2.5)

From standard normal tables, P(Z2.5) = 0.9938

P(Z>2.5)=1-P(Z2.5) = 1 -0.9938=0.0062

probability of getting this result or greater = 0.0062 x 100 = 0.62%

probability of getting this result or greater = 0.62%

e. Assuming that z-scores greater than +2 or less than -2 are unlikely to occur due to sampling error alone, is this a likely or unlikely result

As z-score : 2.5 > +2 ; this is unlikely result.