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According to a Human Resources report, a worker in the industrial countries spends on average 419 minutes a day on the job. Suppose the standard deviation of time spent on the job is 31 minutes.

a. If the distribution of time spent on the job is approximately bell shaped, between what two times would 68% of the figures be?
enter an appropriate value to enter an appropriate value

b. If the distribution of time spent on the job is approximately bell shaped, between what two times would 95% of the figures be?
enter an appropriate value to enter an appropriate value

c. If the distribution of time spent on the job is approximately bell shaped, between what two times would 99.7% of the figures be?
enter an appropriate value to enter an appropriate value

d. If the shape of the distribution of times is unknown, approximately what percentage of the times would be between 359 and 479 minutes?
enter percentages rounded to 1 decimal place % (Round the intermediate values to 3 decimal places. Round your answer to 1 decimal place.)

e. Suppose a worker spent 400 minutes on the job. What would that worker’s z score be, and what would it tell the researcher?
z score = enter the z score rounded to 3 decimal places (Round your answer to 3 decimal places.)

This worker is in the lower half of workers but within select an option standard deviation of the mean.

Sagot :

Answer:

A) Between 388 and 450

B) Between 357 and 481

C) Between 326 and 512

D) 99.995%

E) 72.907%

Step-by-step explanation:

We are given;

Mean; x¯ = 419 minutes

Standard deviation; σ = 31 minutes

A) From the empirical rule, 68% falls within 1 standard deviation.

Thus, it will be between;

419 - 1(31) and 419 + 1(31)

Thus gives;

Between 388 and 450

B) From the empirical rule, 95% falls within 2 standard deviation.

Thus, it will be between;

419 - 2(31) and 419 + 2(31)

Thus gives;

Between 357 and 481

C) From the empirical rule, 99.7% falls within 2 standard deviations.

Thus, it will be between;

419 - 3(31) and 419 + 3(31)

Thus;

Between 326 and 512

D)between 359 and 479 minutes, we will use z-s roe formula;

z = (479 - 359)/31

z = 3.871

From the z-table attached, we see that the probability is 0.99995 and as such, the percentage is 99.995%

E) worker spent 400 minutes on the job.

Thus;

z = (419 - 400)/31

z = 0.613

From z-tables, p = 0.72907

In percentage, we have; 72.907%

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