Question

The 2010 U.S. Census asked every household to report information on each person living there. Suppose for a sample of 30 households selected, the number of persons living in each was reported as follows. 2 3 1 2 6 4 2 1 5 3 2 3 1 2 2 1 3 1 2 2 4 2 1 2 9 3 2 1 1 3 Compute the mean, median, mode, range, lower and upper quartiles, and interquartile range for these data. (Round the mean to 2 decimal places.) Mean

Answer 1

Answer:

$\bar x =2.53$ -- Mean

$Median= 2$

$Mode = 2$

$Range = 8$

$Q_1 = 1$

$Q_3 = 3$

$IQR = 2$

Step-by-step explanation:

Given

n = 30

Data: 2 3 1 2 6 4 2 1 5 3 2 3 1 2 2 1 3 1 2 2 4 2 1 2 9 3 2 1 1 3

Solving (a): The mean

This is calculated as:

$\bar x =\frac{\sum x}{n}$

$\bar x =\frac{2 +3 +1 +2 +6 +4 +2 +1 +5 +3 +2 +3+ 1+ 2 +2 +1 +3 +1 +2 +2 +4 +2 +1 +2 +9 +3 +2 +1 +1 +3}{30}$

$\bar x =\frac{76}{30}$

$\bar x =2.53$

Solving (b): The median

First arrange the given data

Arranged: 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 4 4 5 6 9

$Median =\frac{1}{2}(n+1)$

$Median =\frac{1}{2}(30+1)$

$Median =\frac{1}{2}(31)$

$Median = 15.5th$

This implies that the median is the average of the 15th and 16th item

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

$Median = \frac{4}{2}$

$Median= 2$

Solving (c): The mode

From the given data; 2 has the highest frequency of 11.

So,

$Mode = 2$

Solving (d): The Range

$Range = Highest\ Data - Lowest\ Data$

The highest is 9 and the lowest is 1,

So:

$Range = 9 - 1$

$Range = 8$

Solving (e): Lower (Q1) and Upper (Q3) quartile.

Q1 is calculated as:

$Q_1 = \frac{1}{4}(n+1)th$

$Q_1 = \frac{1}{4}(30+1)th$

$Q_1 = \frac{1}{4}(31)th$

$Q_1 = 7.75th$

$Q_1 = 8th\ item$

$Q_1 = 1$ -- from the arranged data

Q3 is calculated as:

$Q_3 = \frac{3}{4}(n+1)th$

$Q_3 = \frac{3}{4}(30+1)th$

$Q_3 = \frac{3}{4}(31)th$

$Q_3 = 23.25th$

$Q_3 = 23rd\ item$

$Q_3 = 3$ -- from the arranged data

Solving (f): The interquartile range (IQR)

$IQR = Q_3 - Q_1$

$IQR = 3-1$

$IQR = 2$