Question

1. 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 your high school: 7 20 24 25 25 28 28 30 32 35 42 43 44 45 46 47 48 48 50 51 72 75 77 78 79 83 87 88 135 151 (a) construct a box plot or this data. (b) are there any outliers? justify your answer mathematically using the 1.5 * iqr rule. (c) would it be better to use the mean and standard deviation or the median and iqr to describe the spread of this distribution? why?

Explanation:

Step1: Arrange data in ascending order

The data is already in ascending - order: 7, 20, 24, 25, 25, 28, 28, 30, 32, 35, 42, 43, 44, 45, 46, 47, 48, 48, 50, 51, 72, 75, 77, 78, 79, 83, 87, 88, 135, 151.

Step2: Find the median (Q2)

Since \(n = 30\) (an even - numbered data set), \(Q2=\frac{46 + 47}{2}=46.5\).

Step3: Find the lower half and Q1

The lower half of the data is 7, 20, 24, 25, 25, 28, 28, 30, 32, 35, 42, 43, 44, 45, 46. Since \(n_1 = 15\) (an odd - numbered data set), \(Q1 = 28\).

Step4: Find the upper half and Q3

The upper half of the data is 47, 48, 48, 50, 51, 72, 75, 77, 78, 79, 83, 87, 88, 135, 151. Since \(n_2 = 15\) (an odd - numbered data set), \(Q3 = 78\).

Step5: Calculate the inter - quartile range (IQR)

\(IQR=Q3 - Q1=78 - 28 = 50\).

Step6: Calculate the lower and upper fences for outliers

Lower fence \(=Q1-1.5\times IQR=28-1.5\times50=28 - 75=-47\).
Upper fence \(=Q3 + 1.5\times IQR=78+1.5\times50=78 + 75 = 153\).

Step7: Identify outliers

Any value less than the lower fence or greater than the upper fence is an outlier. Since all values in the data set are greater than - 47 and less than 153, there are no outliers.

Step8: Decide on measure of center and spread

Since there are no outliers, the mean and standard deviation can be used to describe the center and spread. The mean is sensitive to extreme values, but since there are no extreme values (outliers) in this data set, the mean and standard deviation will give a good representation of the typical value and the spread around the typical value.

Answer:

(a) To construct a box - plot:

• Draw a number line that includes the range of the data (from 7 to 151).
• Draw a box from \(Q1 = 28\) to \(Q3 = 78\).
• Draw a vertical line inside the box at the median \(Q2 = 46.5\).
• Draw whiskers from the box to the minimum value (7) and the maximum value (151).

(b) There are no outliers. The lower fence is - 47 and the upper fence is 153, and all data values are within this range.
(c) It is better to use the mean and standard deviation. Since there are no outliers, the mean and standard deviation will give a good representation of the center and spread of the data.