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 30 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 the lower limit for the interval where 68% of the values would fall
to enter the upper limit for the interval where 68% of the values would fall

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 the lower limit for the interval where 95% of the values would fall
to enter the upper limit for the interval where 95% of the values would fall

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 the lower limit for the interval where 99.7% of the values would fall
to enter the upper limit for the interval where 99.7% of the values would fall

d. If the shape of the distribution of times is unknown, approximately what percentage of the times would be between 360 and 478 minutes?
At least 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 the distance from the mean
standard deviation of the mean.

We are given the mean and the standard deviation. If the distribution is normal(bell-shaped), the Empirical Rule is used, otherwise, if the distribution is unknown, Chebyshev's Theorem is used.

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

Chebyshev Theorem

The Chebyshev Theorem can also be applied to non-normal distribution. It states that:

At least 75% of the measures are within 2 standard deviations of the mean.

At least 89% of the measures are within 3 standard deviations of the mean.

An in general terms, the percentage of measures within k standard deviations of the mean is given by .

In this question:

Mean of 419, standard deviation of 30.

Question a:

Within 1 standard deviation of the mean, so:

419 - 30 =  389.

419 + 30 = 449.

Between 389 and 449 minutes.

Question b:

Within 2 standard deviations of the mean, so:

419 - 60 = 359

419 + 60 = 479

Between 359 and 479 minutes.

Question c:

Within 3 standard deviations of the mean, so:

419 - 90 = 329

419 + 90 = 509

Between 329 and 509 minutes.

Question d:

Have to find how many standard deviations it is from the mean:

(478 - 419)/30 = 1.97

(360 - 419)/30 = -1.97

Considering

At least 74.2% of the times would be between 360 and 478 minutes.

Question e:

The z-score is:

In which X is the measure, is the mean and is the standard deviation.

For this question, X = 400,

So

This worker is in the lower half of workers but within 0.633 standard deviations of the mean.

For more on the Empirical Rule, you can check link

For more on the Chebyshev Theorem, you can check link

For more on z-scores, you can check link

Step-by-step explanation:

Ah statistics

68-95-99.7

a. 68% is between -1 and 1 standard deviation, which is between 389 and 449 minutes

b. 95% is between -2 and 2 standard deviations, which is between 359 and 479 minutes

c. 99.7% is between -3 and 3 standard deviations, which is between 329 and 509 minutes

d. Use table A with a z-score of (360-419)/30 = -1.96666667

and with a z-score of (478-419)/30 = 1.96666667

NOTE THESE ARE NOT THE ANSWERS, but use the z-scores to find the percentage on table A

e. (400-419)/30 = -0.633333333 as his z-score

now find that value on table A

step-by-step explanation:

goog be able to you with that maybe