Question

The data represents the heights of fourteen basketball players, in inches.

Answer 1

Interquartile range of the old set = 76 - 72 = 4
Interquartile range of the new set = 76 - 73 = 3

IQR of new est < old set.

Answer 2

Answer:

The interquartile range of the new set is less than the interquartile range of the original set.

Step-by-step explanation:

Given : 69, 70, 72, 72, 74, 74, 74, 75, 76, 76, 76, 77, 77, 82

To Find : If the highest and lowest numbers were dropped to form a new set of data how would the inter quartile range of the new set compare to the inter quartile range of the original set?

Solution:

Data : 69, 70, 72, 72, 74, 74, 74, 75, 76, 76, 76, 77, 77, 82

No. of terms = 14

Median = $\frac{\frac{n}{2}\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}$

Median = $\frac{\frac{14}{2}\text{th term}+(\frac{14}{2}+1)\text{th term}}{2}$

Median = $\frac{7\text{th term}+(8)\text{th term}}{2}$

Median = $\frac{74+75}{2}$

Median = $74.5$

$Q_1$ is the median of the lower half of data ( data below the median)

Data : 69, 70, 72, 72, 74, 74, 74

No. of terms = 7

Median = 4th term =72

$Q_1=72$

$Q_3$ is the median of the upper half of data ( data above the median)

Data : 75, 76, 76, 76, 77, 77, 82

No. of terms = 7

Median = 4th term =76

$Q_3=76$

IQR = $Q_3-Q_1=76-72=4$

Now the highest and lowest numbers were dropped to form a new set of data

So, new data : 70, 72, 72, 74, 74, 74, 75, 76, 76, 76, 77, 77

No. of terms = 12

Median = $\frac{\frac{n}{2}\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}$

Median = $\frac{\frac{12}{2}\text{th term}+(\frac{12}{2}+1)\text{th term}}{2}$

Median = $\frac{6\text{th term}+(7)\text{th term}}{2}$

Median = $\frac{74+75}{2}$

Median = $74.5$

$Q_1$ is the median of the lower half of data ( data below the median)

Data: 70, 72, 72, 74, 74, 74,

No. of terms = 6

Median = $\frac{3\text{rd term }+4 \text{th term}}{2}$

Median = $\frac{72+74}{2}=73$

$Q_1=73$

$Q_3$ is the median of the upper half of data ( data above the median)

Data : 75, 76, 76, 76, 77, 77

No. of terms = 6

Median = $\frac{3\text{rd term }+4 \text{th term}}{2}$

Median = $\frac{76+76}{2}=76$

$Q_3=76$

IQR = $Q_3-Q_1=76-73=3$

Thus IQR of old data set is greater than IQR of new data set i.e. 4>3

Hence The interquartile range of the new set is less than the interquartile range of the original set.