Question

Suppose the amount of heating oil used annually by household in Iowa is normally distributed with mean 200 gallons per household per year and a standard deviation of 40 gallons of heating oil per household per year.

(a) If the members of a particular household decided that they wanted to conserve fuel and use less oil than 98% of all other households in Iowa, what is the amount of oil they can use and accomplish their goal?

(b) Suppose a newspaper wants to conduct a survey on heating oil consumption in Iowa. For the survey, the newspaper randomly picks 100 households. What is the mean and standard error of the sampling distribution of the sample mean of heating oil consumption of the 100 households?

(c) Would sample mean of 250 in part (b) cast a doubt on the results of the survey conducted by the newspaper?

Answer 1

Solution:-

a) The amount of oil they can use and accomplish their goal is 117.84.

p-value for the top 98% = 0.02

z-score for the p-value = -2.054

By applying normal distruibution:-

x = 117.84

b) The mean and standard error of the sampling distribution of the sample mean of heating oil consumption of the 100 households would be 200 and 4.00.

S.E = 4.0

c) Yes, the sample mean of 250 in part (b) will cast a doubt on the results of the survey conducted by the newspaper.

x = 250

By applying normal distruibution:-

z = 12.5

P(z > 12.5) = less than 0.001