Answer:
79 because x = 7 so 12 x7 - 5 =79
10 is your answer!! hope this helps
Answer:
Step-by-step explanation:
Hello!
The objective is to estimate the average time a student studies per week.
A sample of 8 students was taken and the time they spent studying in one week was recorded.
4.4, 5.2, 6.4, 6.8, 7.1, 7.3, 8.3, 8.4
n= 8
X[bar]= ∑X/n= 53.9/8= 6.7375 ≅ 6.74
S²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/7*[376.75-(53.9²)/8]= 1.94
S= 1.39
Assuming that the variable "weekly time a student spends studying" has a normal distribution, since the sample is small, the statistic to use to perform the estimation is the student's t, the formula for the interval is:
X[bar] ±
* (S/√n)

6.74 ± 2.365 * (1.36/√8)
[5.6;7.88]
Using a confidence level of 95% you'd expect that the average time a student spends studying per week is contained by the interval [5.6;7.88]
I hope this helps!
Answer:
50%
Step-by-step explanation:
The interval between the first quartile,25-th percentile, (155) and the third quartile, 75-th percentile, (196) comprises 50% of the sample data.
The desired interval is from 155 to 195 pounds, which starts exactly at the first quartile and ends just one pound short of the third quartile. Since this sample has a large number of men and a range of 102 pounds, that one pound difference can be neglected. Therefore, the percent of men who gave a response that falls in the interval 155 to 195 pounds can be guessed to be roughly 50%.