Question

heights of performers (inches) the heights of students (in inches) who are performing in the spring mu - sical are shown in the stem - and - leaf plot. 4|9 5|8 6|1 2 2 3 4 5 5 5 7|2 3 key 6|1 = 61 find the mean, median, mode, range, and interquartile range (iqr) of these heights. press each hotspot. label the corresponding number below with the requested value.

Explanation:

Step1: List out all data values

The data values from the stem - and - leaf plot are 49, 58, 61, 62, 62, 63, 64, 65, 65, 65, 72, 73.

Step2: Calculate the mean

The sum of the data values is \(49 + 58+61 + 62+62+63+64+65+65+65+72+73=759\). There are \(n = 12\) data values. The mean \(\bar{x}=\frac{759}{12}=63.25\).

Step3: Calculate the median

Since \(n = 12\) (an even number), the median is the average of the \(\frac{n}{2}=6\)th and \((\frac{n}{2}+ 1)=7\)th ordered data values. The 6th value is 63 and the 7th value is 64. So the median \(M=\frac{63 + 64}{2}=63.5\).

Step4: Calculate the mode

The mode is the value that appears most frequently. The number 65 appears 3 times, more frequently than any other number, so the mode is 65.

Step5: Calculate the range

The range is the difference between the maximum and minimum values. The maximum value is 73 and the minimum value is 49. So the range \(R=73 - 49 = 24\).

Step6: Calculate the inter - quartile range (IQR)

First, find the lower half and upper half of the data. The lower half is 49, 58, 61, 62, 62, 63 and the upper half is 64, 65, 65, 65, 72, 73.
The median of the lower half (\(Q_1\)) is the average of the 3rd and 4th values. So \(Q_1=\frac{61+62}{2}=61.5\).
The median of the upper half (\(Q_3\)) is the average of the 3rd and 4th values of the upper half. So \(Q_3=\frac{65 + 65}{2}=65\).
The inter - quartile range \(IQR=Q_3 - Q_1=65 - 61.5 = 3.5\).

Answer:

Mean: 63.25
Median: 63.5
Mode: 65
Range: 24
IQR: 3.5