Can you show work or explain how you arrived at the conclusion whenever appropri

Question

Can you show work or explain how you arrived at the conclusion whenever appropriate.

A random sample of 60 employees at a company was obtained. The one-way distance from home to work was recorded for each employee in the sample. Suppose the mean of the sample was 14.8 miles and the standard deviation was 4.1 miles.

a. Compute the standard score (z - score) for a distance of 27 miles. Round your answer to the nearest hundredth and interpret the meaning of your answer as it pertains to this problem.

b. What is the one-way distance from home to work for an employee in the sample who had a standard score (z - score) of -1.50? Round your answer to the nearest tenth.

c. At least what percent of the distances in this sample should we expect to find within 2.5 standard deviations of the mean? Give your answer to the nearest percent.

d. At most what percent of the distances in this sample should we expect to find less than 6.6 miles and more than 23 miles? Give your answer to the nearest percent.

e. At least what percent of the distances in this sample should we expect to find between 8.8 and 20.8 miles? Give your answer to the nearest percent.

3. In a random sample of 645 toner cartridges, the mean number of pages a toner cartridge can print is 4300 and the standard deviation is 345 pages. Assume the distribution of data is normally distributed.

a. Between what two values can you expect to find (the middle) 99.7% of the sample data?

b. What percent of the sample data should you expect to be more than 1 standard deviation away from the mean?

c. How many of the toner cartridges in the sample should you expect to print between 3955 pages and 4645 pages?

d. The company that makes the toner cartridges guarantees to replace any cartridge that prints fewer than 3610 pages. Approximately how many of the cartridges in the sample would you expect to be replaced under the guarantee policy?

Explanation / Answer

We are allowed to do 1 question at a time. Post again for second question.

3) a) n = 645

mean = 4300

SD = 345

a) For the normal distribution, the values within onestandard deviation of the mean account for 68.27% of the set; while within two standard deviations account for 95.45%; and within three standard deviations account for 99.73%.

Between 3 SD

4300 ± 3 * 345

(3265, 5335)

b) 68.27%

c) This is 1 SD away condition

So, 68.27%

Number = 0.6827* 4300 = 2936

d) 3610 is 2SD away

So,

% less than 3610 = 2.275%

Number = 0.02275 * 4300 = 97.825

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.