Answer:
Mean: 40.17 years.
Standard deviation: 10.97 years.
Step-by-step explanation:
The frequency distribution is in the attached image.
We can calculate the mean adding the multiplication of midpoints of each class and frequency, and dividing by the sample size.
The midpoints of a class is calculated as the average of the bounds of the class.
Then, the mean can be written as:

The standard deviation can be calculated as:
![s=\sqrt{\dfrac{1}{N-1}\sum f_i(X_i-E(X))^2}\\\\\\s=\sqrt{\dfrac{1}{59}[3(15-40.17)^2+7(25-40.17)^2+18(35-40.17)^2+20(45-40.17)^2+12(55-40.17)^2]}](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B%5Cdfrac%7B1%7D%7BN-1%7D%5Csum%20f_i%28X_i-E%28X%29%29%5E2%7D%5C%5C%5C%5C%5C%5Cs%3D%5Csqrt%7B%5Cdfrac%7B1%7D%7B59%7D%5B3%2815-40.17%29%5E2%2B7%2825-40.17%29%5E2%2B18%2835-40.17%29%5E2%2B20%2845-40.17%29%5E2%2B12%2855-40.17%29%5E2%5D%7D)

-4(2)+2
-8+2
-6
Hope it helps
Answer:
store C is the best with .122 per ounce the others are .129 and .135
Answer:
w ≤ 593
Step-by-step explanation:
Missing option;
w ≥ 593
w > 593
w ≤ 593
593 < w
Explanation:
w ≥ 593 No
Number of wrapping paper sold never be more than 593.
w > 593 no
Number of wrapping paper sold never be more than 593.
w ≤ 593 yes
Number of wrapping paper sold will be equal or less than 593.