Pulse ratesRefer to Exercise 88.

(a) Find the median pulse rate for all 19 students.

(b) Find the median pulse rate excluding the student with the medical issue. Explain why this student’s 120 pulse rate does not have much effect on the median.

The given pulse rates are:

71, 104, 76, 88, 78, 71, 68, 86, 70, 90, 74, 76, 69, 68, 88, 96, 68, 82, 120

Arrange the given pulse rate from smallest to largest:

68, 68, 68, 69, 70, 71, 71, 74, 76, 76, 78, 82, 86, 88, 88, 90, 96, 104, 120

As we know that median is the middle value of a data set. As the given set is odd, the median will be the middle values (10th) of this sorted data:

$M=76$

therefore, the median is 76 beats per minute.

Exclude the student with a 120 pulse rate as he has medical issues and find the new median.

68, 68, 68, 69, 70, 71, 71, 74, 76, 76, 78, 82, 86, 88, 88, 90, 96, 104

As we know that median is the middle value of a data set. As the given set is even, the median will be the average of the two middle values (9th and 10th) of this sorted data:

$$\begin{gathered} M=\frac{76+76}{2} \\ =\frac{152}{2} \\ =76 \end{gathered}$$

The answer is the same as given in part (a).

Median is a resistant measure of center, the reason is median is not strongly influenced by the extreme observations. Therefore, the resistant property of the median is followed here.