Recent studies indicate that the typical 50-year-old woman spends $353 per year for personal-care products. The distribution of the amounts spent follows a normal distribution with a standard deviation of $44 per year. We select a random sample of 46 women. The mean amount spent for those sampled is $301.

What is the likelihood of finding a sample mean this large or larger from the specified population?(Round z value to 2 decimal places and final answer to 4 decimal places.)

Probability =Answer

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $134,000. This distribution follows the normal distribution with a standard deviation of $30,000.

(a)If we select a random sample of 72 households, what is the standard error of the mean?(Round your answer to the nearest whole number.)

Standard Error =Answer

(b)What is the expected shape of the distribution of the sample mean?Answer

NormalUniformUnknownNot Normal Standard deviation unknown

(c)What is the likelihood of selecting a sample with a mean of at least $137,000?(Round z value to 2 decimal places and final answer to 4 decimal places.)

Probability =Answer

(d)What is the likelihood of selecting a sample with a mean of more than $126,000?(Round z value to 2 decimal places and final answer to 4 decimal places.)

Probability =Answer

(e)

Find the likelihood of selecting a sample with a mean of more than $126,000 but less than $137,000.(Round z value to 2 decimal places and final answer to 4 decimal places.)

Probability =Answer