Chapter 10.4, Problem 23CCR

To determine

To compare: the mean and median of the data set.

Answer to Problem 23CCR

Median of given data set is $\underset{\_}{3\text{stars}}$ while mean of the given data set is $\underset{\_}{2.88\text{stars}}$ .

Explanation of Solution

Given information:

Given data is as follows

Ratings in STARS (x)	Frequency f
4	52
3	85
2	48
1	11
0	2

Formula used :

Mean of dataset = $\frac{{\displaystyle \sum \left(f\times x\right)}}{{\displaystyle \sum f}}$

Median of dataset when sum of frequency that is $n$ is even is value of $x$ where $\frac{n}{2}$ belongs.

Calculation:

Complete the given data set by calculating ${\displaystyle \sum \left(f\times x\right)}$ , ${\displaystyle \sum f}$ and cumulative frequency $n$ .

	Ratings in STARS   $x$	Frequency   $f$	$f\times x$	cumulative frequency   $n$
	4	52	208	52
	3	85	255	52+85=137
	2	48	96	137+48=185
	1	11	11	185+11=196
	0	2	0	196+2=198
Total		198	570

  ${\displaystyle \sum \left(f\times x\right)}=570$

  ${\displaystyle \sum f}=n=198$

Calculate mean using formula $\frac{{\displaystyle \sum \left(f\times x\right)}}{{\displaystyle \sum f}}$

  $mean=\frac{570}{198}=2.8787\approx 2.88$

As $n=198$ is even , $\frac{n}{2}=\frac{198}{2}=99$

As 99 belongs to second row in cumulative frequency having value of $x$ that is ratings in stars as 3 , so median of given dataset is $3$ .

Hence median of given data set is $\underset{\_}{3\text{stars}}$ while mean of the given data set is $\underset{\_}{2.88\text{stars}}$ .