You are interested in how much time students spend on the Internet each day. Here are data on the time spent on the Internet (in minutes) for a particular day reported by a random sample of $30$ students at a large high school:

(a) Construct a histogram of these data.

(b) Are there any outliers? Justify your answer.

(c) Would it be better to use the mean and standard deviation or the median and IQR to describe the center and spread of this distribution? Why?

The figure is

A histogram is a graphing tool that is often used. It's used to summarise data that's either discrete or continuous and measured on an interval scale. It's commonly used to demonstrate the key aspects of data distribution in a user-friendly manner.

For the given data, the proper histogram is as above.

It can be seen from the above histogram that,

1. The minimum: $19$

2. (The median of the initial group is the lower or initial quartile) $Q_1:32$

3. The median: $48$

4. (The median of the second group is the upper or third quartile) $Q_3:80$

5. The maximum: $325$

The mean of the given data is$\overline{x}=\frac{1}{n}\sum _{i=1}^{n}xi$

$n$= a number of values.

$xi,i=1,n$are the values.

Where, If $30$points are given in this case, then $n=30$

The sum of the point is,$\sum _{i=1}^{n}xi=2382$

Thus, the mean$x=\left(\frac{1}{30}\right)2382=\frac{397}{5}=79.4$

The sample standard variation of given data is, s=$\sqrt{\frac{1}{n-1}\sum _{i=1}^n}({xi-x)}^2$

Where,

$n$= a number of values.

$xi,i=1,n$are the values.

$\overline{x}$=mean of the values

The mean of data is,$\overline{x}=\frac{397}{5}$

As here in this example, $30$points are given, then $n=30$

Sum of$$\begin{gathered} {(xi-\overline{x})}^2\sum _{i=1}^n{(xi-\overline{x})}^2=828136/5 \\ {(xi-\overline{x})}^2\sum _{i=1}^n{(xi-\overline{x})}^2=[1/(30-1\left)\right](828136/5) \\ =(1/29)(828136/5) \\ =828136/145 \end{gathered}$$

Finally, s=$\sqrt{828136/145}=\left(2\sqrt{30019930}\right)/145\approx 75.57302930$

Thus, the sample standard deviation is$75.57302930$