Question

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 28 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 391 to enter the upper limit for the interval where 68% of the values would fall 447 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 363 to enter the upper limit for the interval where 95% of the values would fall 475 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 335 to enter the upper limit for the interval where 99.7% of the values would fall 503 d. If the shape of the distribution of times is unknown, approximately what percentage of the times would be between 360 and 478 minutes? enter percentages rounded to 1 decimal place 76.7 % (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 -0.679 (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.

Answer 1

If the distribution is a bell shaped one, then we can assume it to be normal. Therefore

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 391 to enter the upper limit for the interval where 68% of the values would fall 447

Since we are assuming that the population is normally distributed we can use Chebychev's 68-95-99.7 rule.

The first rule is that 68% of data falls within 1 SD of the mean

that is

P( < X < ) = 68%

Therefore the range will be

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 363 to enter the upper limit for the interval where 95% of the values would fall 475

The2nd rule is that 95% of data falls within 2 SD of the mean

that is

P( < X < ) = 95%

Therefore the range will be

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 335 to enter the upper limit for the interval where 99.7% of the values would fall 503

The 3rd rule is that 99.7% of data falls within 3 SD of the mean

that is

P( < X < ) = 99.7%

Therefore the range will be

​​​​​​​

d. If the shape of the distribution of times is unknown, approximately what percentage of the times would be between 360 and 478 minutes? enter percentages rounded to 1 decimal place 76.7 % (Round the intermediate values to 3 decimal places. Round your answer to 1 decimal place.)

P( 360 <X < 478) = P( X < 478) - P( X < 360)

= P( Z < 2.11) - [ 1- P( Z < 2.11) ]

= 2 P( Z < 2.11) - 1

= 2 * 0.98245 - 1 ...................using normal distribution tables

P(360 <X < 478) = 0.9649

= 96.49%

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 -0.679 (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

x = 400

z-score =

z-score basically tells how many SD 'x' is away from the mean.

-1SD < X < 1SD

Therefore this

This worker is in the lower half of workers but within 0.679 standard deviation of the mean.

it is said that he worked for 400 minutes , so within a closest range of 399 - 401.