Question

1. 17) In a certain city, the household electricity bills are approximately normally distributed with a mean of $150 and a standard deviation of $25. 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  $162, 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

Solution :

Given that,

mean = = 150

standard deviation = = 25

n = 25

a) = = 150

b) = / n = 25 / 25 = 5

c) = 162

z = -   /

z = 162 - 150 / 5

z = 2.40

d) P( z > 2.40)

= 1 - P(z < 2.40)   

= 1 - 0.9918

= 0.0082

percentage = 0.82%

e) unlikely, because z-score is greater than +2