Question

The hourly earnings (in dollars) for a sample of 25 railroad equipment manufacturers are:15.60 18.7514.60 15.8014.3513.90 17.5017.5513.8014.20 19.05 15.35 15.20 19.45 15.95 16.50 16.30 15.2515.05 19.10 15.20 16.22 17.75 18.40 15.25Find the median and the mode(s)(if they exist) of the data. What is the interquartile range

Answer 1

Answer:

$Median = 15.80$

$Mode = 15.20\ \&\ 15.25$

$IQR = 2.35$

Step-by-step explanation:

Given

$15.60,\ 18.75,\ 14.60,\ 15.80,\ 14.35,$

$13.90,\ 17.50,\ 17.55,\ 13.80,\ 14.20,$

$19.05,\ 15.35,\ 15.20,\ 19.45,\ 15.95,$

$16.50,\ 16.30,\ 15.25,\ 15.05,\ 19.10,$

$15.20,\ 16.22,\ 17.75,\ 18.40,\ 15.25.$

Solving (a): The median and the mode

First, we sort the data.

$13.80,\ 13.90,\ 14.20,\ 14.35,\ 14.60,\ 15.05,\ 15.20,\ 15.20,\ 15.25,\ 15.25,$

$15.35,\ 15.60,\ 15.80,\ 15.95,\ 16.22,\ 16.30,\ 16.50,\ 17.50,\ 17.55,\ 17.75,$

$18.40,\ 18.75,\ 19.05,\ 19.10,\ 19.45.$

The median position is:

$Median = \frac{n + 1}{2}th$

$Median = \frac{25 + 1}{2}$

$Median = \frac{26}{2}$

$Median = 13th$

The 13th item is: 15.80

Hence:

$Median = 15.80$

The modes are:

$Mode = 15.20\ \&\ 15.25$ --- they both have frequency of 2 while others occur once

Solving (b): The interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

Since the median is at the 13th position, Q1 is:

$Q_1 = \frac{1 + 13}{2}th$

$Q_1 = \frac{14}{2}th$

$Q_1 = 7th$

The 7th item is: 15.20

$Q_1 = 15.20$

Similarly, Q3 is:

$Q_3 = \frac{13+n}{2}$

$Q_3 = \frac{13+25}{2}$

$Q_3 = \frac{38}{2}$

$Q_3 = 19th$

The 7th item is: 17.55

So:

$Q_3 = 17.55$

Hence,

$IQR = 17.55 - 15.20$

$IQR = 2.35$