Question

SCCoast, an Internet provider in the Southeast, developed the following frequency distribution on the age of Internet users. Find the mean and the standard deviation. (Round squared deviations to nearest whole number and final answer to 2 decimal places.)

Answer 1

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:

$E(X)=\dfrac{1}{N}\sum f_iX_i\\\\\\E(X)=\dfrac{1}{60}(3\cdot15+7\cdot25+18\cdot35+20\cdot45+12\cdot 55)=\dfrac{45+175+630+900+660}{60}\\\\\\E(X)=\dfrac{2,410}{60}=40.17$

The standard deviation can be calculated as:

$s=\sqrt{\dfrac{1}{N-1}\sum f_i(X_i-E(X))^2}\\\\\\s=\sqrt{\dfrac{1}{59}[3(15-40.17)^2+7(25-40.17)^2+18(35-40.17)^2+20(45-40.17)^2+12(55-40.17)^2]}$

$\\\\\\s=\sqrt{\dfrac{1}{59}( 1,900.59 + 1,610.90 + 481.12 + 466.58 + 2,639.15 )}\\\\\\s=\sqrt{\dfrac{ 7,098.33 }{59}}=\sqrt{120}=10.97$