Question

source: world almanac 2012

1. executive bonuses a random sample of bonuses (in millions of dollars) paid by large companies to their executives is shown. find the mean and modal class for the data.

class boundaries frequency
0.5 - 3.5 11
3.5 - 6.5 12
6.5 - 9.5 4
9.5 - 12.5 2
12.5 - 15.5 1

Explanation:

Step1: Find the mid - points of each class

For the class $0.5 - 3.5$, the mid - point $x_1=\frac{0.5 + 3.5}{2}=2$. For $3.5 - 6.5$, $x_2=\frac{3.5+6.5}{2}=5$. For $6.5 - 9.5$, $x_3=\frac{6.5 + 9.5}{2}=8$. For $9.5 - 12.5$, $x_4=\frac{9.5+12.5}{2}=11$. For $12.5 - 15.5$, $x_5=\frac{12.5 + 15.5}{2}=14$.

Step2: Calculate the product of mid - points and frequencies

Let $f_1 = 11,f_2=12,f_3 = 4,f_4=2,f_5 = 1$. Then $f_1x_1=11\times2 = 22$, $f_2x_2=12\times5=60$, $f_3x_3=4\times8 = 32$, $f_4x_4=2\times11=22$, $f_5x_5=1\times14 = 14$.

Step3: Calculate the sum of frequencies and the sum of $f\times x$

$\sum f=f_1 + f_2+f_3+f_4+f_5=11 + 12+4+2+1=30$. $\sum(fx)=22+60+32+22+14 = 150$.

Step4: Calculate the mean

The mean $\bar{x}=\frac{\sum(fx)}{\sum f}=\frac{150}{30}=5$.

Step5: Find the modal class

The class with the highest frequency is $3.5 - 6.5$ (since the frequency $12$ is the highest among all frequencies).

Answer:

Mean: $5$; Modal class: $3.5 - 6.5$