Question

The time taken to prepare the envelopes to mail a weekly report to all executives in a company has a normal distribution, with a mean of 35 minutes and a standard deviation of 2 minutes. On 95% of occasions the mailing preparation takes less than a) 38.29 minutes b) 31.71 minutes c) 35.25 minutes d) 34.75 minutes

Answer 1

Answer:

Mailing preparation takes 38.29 min max time to prepare the mails.

Step-by-step explanation:

Given:

Mean:35 min

standard deviation:2 min

and 95%  confidence interval.

To Find:

In normal distribution mailing preparation time  taken less than.

i.eP(t<x)=?

Solution:

Here t -time and x -required time

mean time 35 min

5 % will not have true mean value . with 95 % confidence.

Question is asked as ,preparation takes less than  time means what is max time that preparation will take to prepare mails.

No mail take more time than that time .

by Z-score or by confidence interval is

Z=(X-mean)/standard deviation.

Z=1.96 at  95 % confidence interval.

1.96=(X-35)/2

3.92=(x-35)

X=38.29 min

or

Confidence interval =35±Z*standard deviation

=35±1.96*2

=35±3.92

=38.29 or 31.71 min

But we require the max time i.e 38.29 min

And by observation we can also conclude the max time from options as 38.29 min.