According to a Human Resources report, a worker in the industrial countries spen

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 29 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? 392 446 to b. If the distribution of time spent on the job is approximately bell shaped, between what two times would 95% of the figures be? 365 473 to 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? 338 500 to d. If the shape of the distribution of times is unknown, approximately what percentage of the times would be between 360 and 478 minutes? 97.36 % (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? 24.20(Roun z score = Round your answer to 3 decimal places.) two + This worker is in the lower half of workers but within standard deviation of the mean

Explanation / Answer

a)as 1 std deviation away values have 68% between them

therefore interval =mean -/+1*std deviation =419 -/+ 29 =390 to 448

b)as 2 std deviation away values have 68% between them

therefore interval =mean -/+ 2*std deviation =419 -/+ 2*29 =361 to 477

c)

as 3 std deviation away values have 99.7% between them

therefore interval =mean -/+ 3*std deviation =419 -/+ 3*29 =332 to 506

d)for distribution is unknown from Chebychev's

P(360<X<478)=P((360-419)/29<Z<(478-419)/29)=P(-2.0345<Z<2.0345)=(1-1/2.03452)*100 =75.8%

e)

z score =(X-mean)/std deviation =(400-419)/29 =-0.655

this worker is the lower half of workers but within 1 std deviation of the mean

( please revert for any clarification required)

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