Question

You are interested in how many contacts older adults have in their smartphones. Here are data on the number of contacts for a random sample of 30 elderly adults with smartphones in a large city:

Part (a). Construct a histogram of these data.

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

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

Short answer

Part (a)

Part (b)

151 is an outlier
Part (c) Median and IQR

Step-by-step solution

 Part (a) Step 1. Given information.

Here are the contact numbers for a random sample of 30 elderly adults in a large city who own smartphones:

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

 Part (a) Step 2.Create a histogram from these data.

Let's start by calculating the frequency of each interval, which is the number of months of data values that fall within each interval. The first interval begins at Handle and has a width of 20, so it is 0-20. The next intervals will be 20-40, 40-60, and so on until all data values are assigned to exactly one interval.

Interval	Frequency
0<20	1
20<40	9
40<60	10
60<80	5
80<100	3
100<120	0
120<140	1
140<160	1

Histogram of Frequency

Each bar must have the same width and be bounded by the interval boundaries, while the height must be equal to the frequency.

 Part (b) Step 1.There any outliers 

Let's start by determining all data values from the dotplot, where the data values correspond to the dots in the dotplot.

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

The data values in the dotplot were already ordered.

The median is the sorted data set's middle value. Because the number of data values is even, the median is the average of the sorted data set's two middle values (15th and 16th data values):

$$\begin{gathered} M=Q_2=\frac{46+47}{2} \\ \frac{46+47}{2}=\frac{93}{2} \\ 46.5 \end{gathered}$$

The first quartile is the median of the data values that are less than the median (or at 25 percent of the data). Because there are 15 data points below the median, the first quartile is the eighth data point.

$Q_1=30$

The third quartile is the median of the data values that are greater than the median (or at 75 percent of the data). The third quartile is the 23rd data value because there are 15 data values above the median.

$Q_3=77$

The IQR is the difference between the third and first quartiles:

$$\begin{gathered} IQR=Q_3-Q_3 \\ {Q}_3-Q_3=77-30 \\ 77-30=47 \end{gathered}$$

The IQR is the difference between the third and first quartiles:

$$\begin{gathered} Q_3+1.5IQR=77+1.5\left(47\right) \\ 77+1.5\left(47\right)=147.5 \\ Q1-1.5IQR=30-1.5\left(47\right)=-40.5 \\ 30-1.5\left(47\right)=-40.5 \end{gathered}$$

We then observe that 151 is an outlier because it is greater than 147.5.

As a result:

151 is an outlier

 Part (c) Step 1. Would it be better to use the mean and standard deviation or the median and IQR to describe the center and variability of this distribution 

Part (b) informs us that the data set contains an outlier.

The mean and standard deviation are not resistant, but the median and IQR are. This means that outliers have a strong influence on the mean and standard deviation, but not on the median and IQR.

In this case, the median and IQR are more reliable because they are less influenced by the outlier in the data set. As a result, the median and IQR are the best options.

As a result:

Median and IQR