A data set includes 104body temperatures of healthy adult humans having a mean of 98.7degreesFand a standard deviation of 0.66degreesF. Construct a 99%confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6degreesFas the mean body temperature?

The 99% confidence interval estimate of the mean body temperature of all healthy humans is between 98.53ºF and 98.87 ºF. 98.6ºF is part of the interval, so the sample suggests that the use of 98.6ºF as the mean body temperature is correct.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 104 - 1 = 103

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 103 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.624

The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 98.7 - 0.17 = 98.53ºF.

The upper end of the interval is the sample mean added to M. So it is 98.7 + 0.17 = 98.87 ºF.

The 99% confidence interval estimate of the mean body temperature of all healthy humans is between 98.53ºF and 98.87 ºF. 98.6ºF is part of the interval, so the sample suggests that the use of 98.6ºF as the mean body temperature is correct.

62.5 paisa

step-by-step explanation:

4000*70 = 280000

280000-(2000*90) = 100000

4000-2400 = 1600

100000/1600 = 62.5

( give brainliest)

explanation:

step-by-step explanation: