Question

1. hourly compensation for production workers the hourly compensation costs (in u.s. dollars) for production workers in selected countries are represented below. find the mean and modal class for the data.

class frequency
2.48 - 7.48 7
7.49 - 12.49 3
12.50 - 17.50 1
17.51 - 22.51 7
22.52 - 27.52 5
27.53 - 32.53 5
compare the mean of these grouped data to the u.s. mean of $21.97.
source: new york times almanac

Explanation:

Step1: Find the mid - points of each class

For the class $2.48 - 7.48$, the mid - point $x_1=\frac{2.48 + 7.48}{2}=4.98$.
For the class $7.49 - 12.49$, the mid - point $x_2=\frac{7.49+12.49}{2}=9.99$.
For the class $12.50 - 17.50$, the mid - point $x_3=\frac{12.50 + 17.50}{2}=15.00$.
For the class $17.51 - 22.51$, the mid - point $x_4=\frac{17.51+22.51}{2}=20.01$.
For the class $22.52 - 27.52$, the mid - point $x_5=\frac{22.52+27.52}{2}=25.02$.
For the class $27.53 - 32.53$, the mid - point $x_6=\frac{27.53+32.53}{2}=30.03$.

Step2: Calculate the product of mid - points and frequencies

$f_1x_1=7\times4.98 = 34.86$.
$f_2x_2=3\times9.99 = 29.97$.
$f_3x_3=1\times15.00 = 15.00$.
$f_4x_4=7\times20.01 = 140.07$.
$f_5x_5=5\times25.02 = 125.10$.
$f_6x_6=5\times30.03 = 150.15$.

Step3: Calculate the sum of frequencies and the sum of products

$\sum f_i=7 + 3+1+7+5+5=28$.
$\sum f_ix_i=34.86+29.97 + 15.00+140.07+125.10+150.15=495.15$.

Step4: Calculate the mean

The mean $\bar{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{495.15}{28}\approx17.68$.

Step5: Find the modal class

The class with the highest frequency is $2.48 - 7.48$ and $17.51 - 22.51$ (both have a frequency of 7). So the modal classes are $2.48 - 7.48$ and $17.51 - 22.51$.

Answer:

Mean: $\approx17.68$
Modal class: $2.48 - 7.48$ and $17.51 - 22.51$