Asimple random sample from a population with a normal distribution of 106106 body temperatures has x overbarxequals=98.1098.10degrees upper f°f and sequals=0.610.61degrees upper f°f. construct anan 8080% confidence interval estimate of the standard deviation of body temperature of all healthy humans.

Using the chi-square distribution, the 80% confidence interval estimate of the standard deviation of body temperature of all healthy humans is (0.5477, 0.6856).

The sample size is

The significance level is

The estimate, which is the sample standard deviation, is of .

Now, we have to find the critical values for the chi-square distribution. They are:

The confidence interval for the population variance is:

The standard deviation is the square root of the variance, thus:

.

The 80% confidence interval estimate of the standard deviation of body temperature of all healthy humans is (0.5477, 0.6856).

A similar problem is given at link

(97.084, 99.116)

Step-by-step explanation:

As per given , we have

sample size :n= 106

degree of freedom: df = 105      (df=n-1)

Significance interval :

Since , population standard deviation is unknown , so we use t-test .

Using t-value table ,

Now , 80% Confidence interval will be :

Thus , The 80% confidence interval estimate of the standard deviation of body temperature of all healthy humans = (97.084, 99.116).

the answer is 8

step-by-step explanation:

3.14 x 8x8

sorry can't

step-by-step explanation:

i googled though and something came up but it was in science langueage not common english sooo