Question

Data from the Bureau of Labor Statistics' Consumer Expenditure Survey show customers spend an average of (µ) $608 a year for cellular phone services. The standard deviation of annual cellular spending is (σ) $132. The random variable, yearly cellular spending, is denoted by X. We plan to select a random sample of 100 cellular customers.

11.

The sampling distribution of X¯X¯

Select one:

a. is not normal because the sample size is too small

b. is normal due to the Chebyshev's Theorem

c. is normal due to the Central Limit Theorem

d. is not normal because the sample size is too large

12.

The standard error (SE) of X¯X¯ is

Select one:

a. 60.8

b. 1.32

c. 132

d. 13.2

13.

What is the probability that a random sample of 100 cellular customers will provide an average(X¯X¯) that is within $25 of the population mean (µ)?

Select one:

a. 3%

b. 94%

c. 97%

d. 6%

14.

The probability in the PRECEDING question would ------ if we were to increase the sample size (n) from 100 to 151.

Select one:

a. be zero

b. stays the same

c. increase

d. decrease

15.

Suppose we reduce the sample size (n) from 100 to 25. The sampling distribution of X¯X¯ will be normal only if

Select one:

a. X has a right skewed distribution

b. X is normally distributed

c. X has a bi-modal distribution

d. X has a left skewed distribution

Answer 1

Solution-:

Let, X- custmer yearly cellular spending

Given:

11) The sampling distribution of is,

Option: c)  is normal due to the Central Limit Theorem

12) The standard error (SE) of is,

  

Option: d) 13.2 is correct

13) we find ,

P[a random sample of 100 cellular customers will provide an average that is within $25 of the population mean]

  

  

  

   (From Z table)

  

The required answer is 0.97*100=97%

Option: c) 97% is correct

14) If we were to increase the sample size (n) from 100 to 151 in queation (13) then

P[a random sample of 151cellular customers will provide an average that is within $25 of the population mean]

  

  

  

   (From Z table)

  

The required aprobability is 0.9990

Option:c. Increase is correct

15) Suppose we reduce the sample size (n) from 100 to 25. The sampling distribution of will be normal only if

Option: b. X is normally distributed