Chapter 10.3, Problem 8PPS

To determine

To calculate mean and median and state the effect of removing outlier

Answer to Problem 8PPS

Mean is 96.18 and Median is 95. Mean would be more affected by removal of outliers.

Explanation of Solution

Given information:

Daily attendance
86	45	91	104	95	88
127	85	79	102	98	103

Formula used: Mean is,

Mean= $\frac{{\displaystyle \sum _{i=1}^n{X}_i}}{n}$

Where, n=total number of terms,

Median is,

When n is odd, $median={(\frac{N+1}{2})}^{th}ob$

When n is even, $median=95$

Calculation:

Re-arranging the data in an ascending order.

45,79,85,86,88,91,95,98,102,103,104,127

For mean substituting the values,

  $mean=\frac{\text{45+79+85+86+88+91+95+98+102+103+104+127}}{12}$

So,

  $mean=\frac{1103}{12}$

On solving,

  $mean=91.92$

Hence, Mean is 91.92

For median substituting the value,

  $median=\frac{{(\frac{12}{2})}^{th}+{(\frac{12}{2}+1)}^{th}}{2}ob$

On solving,

  $median=\frac{{(6)}^{th}+{(7)}^{th}}{2}ob$

Substituting the value of 6^th and 7^th observation,

  $median=\frac{95+98}{2}$

On solving,

  $median=96.5$

Hence, median is 96.5

The outlier is 45.

After removing outlier, new data points are

79,85,86,88,91,95,98,102,103,104,127

For new mean, Substituting values in the mean formula,

= $\frac{\text{79+85+86+88+91+95+98+102+103+104+127}}{11}$

So,

  $=\frac{1058}{2}$

On solving,

  $=96.18$

Hence, new mean is 96.18

For new median, substituting values,

= ${(\frac{11+1}{2})}^{th}ob$

So,

= ${(\frac{12}{2})}^{th}ob$

On solving,

= $6^{th}ob$

On substituting the value of 6^th observation,

  $median=95$

Comparing the new and old mean.

By removing the outliers, the new mean is increased. But, the median is decreased when the outlier is removed.