The ideal (daytime) noise-level for hospitals is 45 decibels with a standard deviation of 10 db; which is to say, this may not be true. A simple random sample of 75 hospitals at a moment during the day gives a mean noise level of 47 db. Assume that the standard deviation of noise level for all hospitals is really 10 db. All answers to two places after the decimal. g

(a) A 99% confidence interval for the actual mean noise level in hospitals is (44.02 db, 49.98 db).

(b) We can be 90% confident that the actual mean noise level in hospitals is 47 db with a margin of error of 1.89 db.

(c) Unless our sample (of 81 hospitals) is among the most unusual 2% of samples, the actual mean noise level in hospitals is between 44.41 db and 49.59 db.

Step-by-step explanation:

The problem is incomplete. The questions are:

(a) A 99% confidence interval for the actual mean noise level in hospitals is (44.02 db, 49.98 db).

For a 99% CI, the value of z is z=2.58

Then, the confidence interval for the mean is:

(b) We can be 90% confident that the actual mean noise level in hospitals is 47 db with a margin of error of 1.89 db.

For a 90% CI, the value of z is z=1.64.

Then, we can calculate the margin of error as:

(c) Unless our sample (of 81 hospitals) is among the most unusual 2% of samples, the actual mean noise level in hospitals is between 44.41 db and 49.59 db.

The 2% tails data corresponds, in the standard normal distirbution, to the values of z whose absolute value is higher than 2.33.

The values of db for these critical values are:

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation:

answer

106 emergencies

step-by-step explanation: