Question

the attendance at college of a group of 18 students was recorded for a 4 week period.
the numbers of students actually attending each of 16 classes are shown below.
18,16,18,15,18,17,14,17,17,16,17,18,17,18,18,16.calculate the mean and standard deviation of students attending the classes.
EXPRESS THE MEAN AS A PERCENTAGE OF THE 18 STUDENTS IN THE GROUP.

Answer 1

Conditions

the attendance at college of a group of 18 students was recorded for a 4 week period.

the numbers of students actually attending each of 16 classes are shown below.

18,16,18,15,18,17,14,17,17,16,17,18,17,18,18,16 calculate the mean and standard deviation of students attending the classes.

EXPRESS THE MEAN AS A PERCENTAGE OF THE 18 STUDENTS IN THE GROUP.

Solution

For calculating the mean we must use the formula of expectation for a discrete random value:

$$M (X) = \frac {1}{n} \sum_ {i = 1} ^ {n} x _ {i}$$

where $n = 16$ , $x_{i}$ - the number of students actually attending each of $n = 16$ classes from 1st to 16th.

$$\begin{array}{l} M (X) = \frac {1}{1 6} (1 8 + 1 6 + 1 8 + 1 5 + 1 8 + 1 7 + 1 4 + 1 7 + 1 7 + 1 6 + 1 7 + 1 8 + 1 7 + 1 8 + \\ 1 8 + 1 6) = 1 6, 8 7 5 \\ \end{array}$$

To find a standard deviation we must use the formula below:

$$\sigma = \sqrt {\frac {1}{n} \sum_ {i = 1} ^ {n} \left(x _ {i} - M (X)\right) ^ {2}}$$

These calculations are comfortably to do in Excel (here is a table):

The standard deviation is 1,165922382. And this is $\frac{1,165922382}{18} \cdot 100\% \approx 6,477347\%$ of all group.

Answer:

The mean is $M(X) = 16,875$

The standard deviation is $\sigma = 1,165922382 = 6,477347\%$