multiple 72 by .05 (36) to find 10 percent, then multiple 80 percent by that number(36) then add both numbers together to get total number of members that month

(72x.05)=36x2+(.80x36)=288

288+72=360

3 0

3 years ago

The rule for calculating the mean is to add up all the scores in a sample and divide by the _________

Tems11 [23]

Amount of numbers in the sample.

8 0

3 years ago

On day two of a study on body temperatures, 106 temperatures were taken. Suppose that we only have the first 10 temperatures to

NemiM [27]

Answer:

The 95% confidence interval for the mean of all body temperatures is between 97.76 ºF and 99.12 ºF

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 = 10 - 1 = 9

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 2.2622

The margin of error is:

M = T*s = 2.2622*0.3 = 0.68

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 98.44 - 0.68 = 97.76 ºF

The upper end of the interval is the sample mean added to M. So it is 98.44 + 0.68 = 99.12 ºF

The 95% confidence interval for the mean of all body temperatures is between 97.76 ºF and 99.12 ºF

4 0

2 years ago