Question

The answer to the question. And what is the variance of the ages?

Answer 1

Let the age of students in a certain public high school be x;

So, we create a table of values as follows;

First, we fine the mean of the distribution, we have;

`\begin{gathered} \mu=\sum ^{}_{}x\cdot P(x) \\ \mu=1.04+3.64+3.45+4.16+2.38+0.54 \\ \mu=15.21 \end{gathered}`

Thus, the standard deviation of the distribution is;

`\sigma=\sqrt[]{\sum^{}_{}(x_i-\mu)^2* P(x_i)}`

Then, we will insert the deviation in the table above,

Then, we have;

`\sum ^{}_{}(x_i-\mu)^2* P(x_i)=1.6259`
`\begin{gathered} \sigma=\sqrt[]{1.6259} \\ \sigma=1.2751 \end{gathered}`

The standard deviation of the ages is 1.2751

Also, the variance is;

The variance is the square of standard deviation, we have;

`\begin{gathered} \sigma^2=(1.2751)^2 \\ \sigma^2=1.6259 \end{gathered}`

Thus, the variance of the ages is 1.6259