Question

the movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer. which of the following statements about the box - and - whisker plot below is not true? the median of the data set shown is 55. both 25 and 145 are values in the set of data. the midrange of the data set shown is 85. there are more data points in the upper two quartiles than in the lower two quartiles.

Explanation:

Step1: Recall Box - Whisker Plot Concepts

A box - and - whisker plot shows the minimum, first quartile (\(Q_1\)), median (\(Q_2\)), third quartile (\(Q_3\)), and maximum of a data set. Each quartile (including the median) divides the data set into four equal - sized groups (quartiles), so each quartile has approximately \(\frac{1}{4}\) of the data points.

Step2: Analyze Option A (Median)

The median is the line inside the box. From the plot, the box starts at 40, the median line is at 55 (mid - point of 40 and 90? Wait, no, the box is from 40 to 90? Wait, the first part of the box is from 40 to 55? Wait, the box is divided into two parts. The median is the middle line of the box. If the left part of the box is from 40 to 55 and the right part is from 55 to 90, then the median is 55. So the statement "The median of the data set shown is 55" is true.

Step3: Analyze Option B (Data Points)

The whiskers extend to the minimum and maximum values. The left whisker starts at 25 (since the left whisker goes to 25) and the right whisker goes to 145. So 25 (minimum) and 145 (maximum) are values in the data set. So the statement "Both 25 and 145 are values in the set of data" is true.

Step4: Analyze Option C (Midrange)

The midrange is calculated as \(\frac{\text{minimum}+\text{maximum}}{2}\). The minimum is 25 and the maximum is 145. Then midrange \(=\frac{25 + 145}{2}=\frac{170}{2}=85\). So the statement "The midrange of the data set shown is 85" is true.

Step5: Analyze Option D (Quartiles and Data Points)

Since each quartile (the lower two quartiles: from minimum to \(Q_2\) (median), and the upper two quartiles: from \(Q_2\) to maximum) contains approximately \(\frac{1}{2}\) of the data? Wait, no. The lower two quartiles: from minimum to \(Q_2\) (median) consists of the first two quartiles (\(Q_1\) to \(Q_2\) and minimum to \(Q_1\)), so it's half of the data. The upper two quartiles: from \(Q_2\) to \(Q_3\) and \(Q_3\) to maximum, also half of the data. Wait, each quartile ( \(Q_1 - Q_2\), \(Q_2 - Q_3\), minimum - \(Q_1\), \(Q_3 -\) maximum) has \(\frac{1}{4}\) of the data. So the lower two quartiles (minimum - \(Q_2\)) have \(\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\) of the data, and the upper two quartiles (\(Q_2 - \) maximum) also have \(\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\) of the data. So the number of data points in the upper two quartiles is equal to the number of data points in the lower two quartiles. So the statement "There are more data points in the upper two quartiles than in the lower two quartiles" is not true.

Answer:

D. There are more data points in the upper two quartiles than in the lower two quartiles.