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The mean amount of life insurance per household is $117,000. This distribution is positively skewed. The standard deviation of the population is $33,000. UseAppendix B.1for thez-values.

a.A random sample of 70 households revealed a mean of $121,000. What is the standard error of the mean?(Round the final answer to 2 decimal places.)

Standard error of the mean

b.Suppose that you selected 121,000 samples of households. What is the expected shape of the distribution of the sample mean?

Shape

(Click to select)

Uniform

Unknown

Normal

Not normal, the standard deviation is unknown

c.What is the likelihood of selecting a sample with a mean of at least $121,000?(Round the final answer to 4 decimal places.)

Probability

d.What is the likelihood of selecting a sample with a mean of more than $109,000?(Round the final answer to 4 decimal places.)

Probability

e.Find the likelihood of selecting a sample with a mean of more than $109,000 but less than $121,000.(Round the final answer to 4 decimal places.)

Probability

1. 

Check my work The Wachesaw Manufacturing Co. produced the following number of units in the last 16 days: 125 127 134 129 128 126 126 129 127 129 127 129 132 131 127 127 The information is to organized into a frequency distribution. a. How many classes would you recommend? . What class interval would you suggest? (Round the final answer to 1 decimal place.) . What lower limit would you recommend for the first class? d. Organize the information into a frequency distribution and determine the relative frequency distribution. (Round the Ch Relative Units Frequency frequency 125 to under 127.0 127.0 to under 129.0 129.0 to under 131.0 131.0 to under 133.0 133.0 to under 135.0 Total e. Comment on the shape of the distribution. The largest concentration is in the up to frequency (