Chapter 10.3, Problem 2GP

To determine

To calculate mean and median. Identify outliers, if any and there effect when removed.

Answer to Problem 2GP

Mean is 51.5, Median is 42.5. Mean would be decreased by removal of outliers.

Explanation of Solution

Given information: Data set is,

25,35,45,30,65,50,25,100,45,35,5,105,110,190,40,30,80

Formula used: Mean formula is, Mean= $\frac{{\displaystyle \sum _{i=1}^n{X}_i}}{n}$

Where, n=total number of terms,

Median formula is,

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

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

Calculation:

Arranging the data in ascending order,

5, 25, 25, 30, 30, 35, 35, 40, 45, 45, 50, 65, 80, 100, 105, 110, 190

For mean, substituting the values in mean formula,

  $mean=\frac{5+25+25+30+30+35+35+40+45+45+50+65+80+100+105+110+190}{17}$

So,

  $mean=\frac{1015}{17}$

On solving,

  $mean=59.71$

Hence, mean is 59.71.

For medium, substituting the values in median formula,

  $median=\frac{17+1}{2}ob$

So,

  $median=\frac{18}{2}ob$

On solving,

  $median=9^{th}ob$

On substituting the value,

  $median=45$

For outliers,

An outlier is an observation that lies an abnormal distance from other values in a random sample from a population.

Comparing the definition and the data set.

The data set has 2 outliers 5 and 190.

Removing the outliers from the dataset. Resulting data set is,

25, 25, 30, 30, 35, 35, 40, 45, 45, 50, 65, 80, 100, 105, 110

For mean, substituting the new values,

  $mean=\frac{25+25+30+30+35+35+40+45+45+50+65+80+100+105+110}{16}$

So,

  $mean=\frac{820}{16}$

On solving,

  $mean=51.25$

Hence, the new mean is 51.25.

For median, substituting the new values,

  $median=\frac{(\frac{{16}^{th}}{2})ob+{(\frac{16}{2}+1)}^{th}ob}{2}$

On solving,

  $median=\frac{8^{th}ob+9^{th}ob}{2}$

Substituting the values of 8^th and 9^th observation,

  $median=\frac{45+45}{2}$

On solving,

  $median=45$

Hence, the new median is 45.

Comparing the new and old mean and median.

The removal of outliers decreased the value of mean by 8.48 units, but, the removal of outliers have no effect on the median of the dataset,