Feeling sleepy? The first four students to arrive for a first-period statistics class were asked how much sleep (to the nearest hour) they got last night. Their responses were $7,7,9,$and $9$

(a) Find the standard deviation from its definition. That is, find the deviations of each observation from the mean, square the deviations, then obtain the variance and the standard deviation.

(b) Interpret the value of $s_x$ you obtained in (a).

(c) Do you think it’s safe to conclude that the mean amount of sleep for all $30$ students in this class is close to $8$ hours? Why or why not?

The data values are $7,7,9,9$

Calculate the average first.

Step 2: Determine each score's standard deviation.

Step 3: Calculate the square root of each standard deviation.

Step 4: Add up the squares to get the total.

Step 5: Work out how much of a difference there is.

Step 6: Determine the square root of the variance.

The number of values in the sample size is: $n=4$

The mean is calculated by dividing the total number of values by the sample size: $\overline{X}=\frac{7+7+9+9}{4}=\frac{32}{4}=8$

If $Xi$are the data values, then $Xi-\overline{X}$are the deviations,${(Xi-\overline{X})}^2$

the squared deviations:

The variance is calculated by dividing the sum of the squared deviations by n-1: ${s_x}^2=\frac{1+1+1+1}{4-1}=\frac{4}{3}$

The square root of the variance is the standard deviation: $s_x=\sqrt{\frac{4}{3}}\approx 1.1547$

Hence, $1.1547$ is the standard deviation.

This suggests that the difference between a typical amount of sleep and the mean should be $1.1547$ hours. Hence, $1.1547$ is the expected distance.

No, because the students were not chosen at random and the sample was a convenience sample, the answer is no. The results of a convenience sample cannot be applied to the entire population. The students that arrive earliest to class, for example, are likely to require the least amount of sleep.