The ages of a random sample of five university professors are 39, 54, 61, 72, and 59. Using this information, find a 99% confidence interval for the population standard deviation of the ages of all professors at the university, assuming that the ages of university professors are normally distributed

99% confidence interval for the population standard deviation = (74.97 , 635.20).

Step-by-step explanation:

We are given that the ages of a random sample of five university professors are 39, 54, 61, 72 and 59. Also, it is provided that the ages of university professors are normally distributed.

So, firstly the pivotal quantity for 99% confidence interval for the population standard deviation is given by;

         P.Q. = ~

where, s = sample standard deviation

            = population standard deviation

            n = sample of university professors = 5

Also, = = 144.5

So, 99% confidence interval for population standard deviation, is;

P(0.2070 < < 14.86) = 0.99 {As the table of at 4 degree of freedom

                                                      gives critical values of 0.2070 & 14.86}

P(0.2070 < < 14.86) = 0.99

P( < < ) = 0.99

P( < < ) = 0.99

99% confidence interval for = ( , )

                                                   = ( , )

                                                   = (5620.525 , 403483.092)

99% confidence interval for = ( , )

                                                  = (74.97 , 635.20)

Therefore, 99% confidence interval for the population standard deviation of the ages of all professors at the university is (74.97 , 635.20).

d. 8

step-by-step explanation:

the given radical expression is:

we can rewrite this as:

we express as square numbers to get;

the exponents will now cancel out to give.

the correct answer is d

d. all of the above

step-by-step explanation: