Answer:
[tex]\bar x =2.53[/tex] -- Mean
[tex]Median= 2[/tex]
[tex]Mode = 2[/tex]
[tex]Range = 8[/tex]
[tex]Q_1 = 1[/tex]
[tex]Q_3 = 3[/tex]
[tex]IQR = 2[/tex]
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:
[tex]\bar x =\frac{\sum x}{n}[/tex]
[tex]\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}[/tex]
[tex]\bar x =\frac{76}{30}[/tex]
[tex]\bar x =2.53[/tex]
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
[tex]Median =\frac{1}{2}(n+1)[/tex]
[tex]Median =\frac{1}{2}(30+1)[/tex]
[tex]Median =\frac{1}{2}(31)[/tex]
[tex]Median = 15.5th[/tex]
This implies that the median is the average of the 15th and 16th item
[tex]Median = \frac{2+2}{2}[/tex]
[tex]Median = \frac{4}{2}[/tex]
[tex]Median= 2[/tex]
Solving (c): The mode
From the given data; 2 has the highest frequency of 11.
So,
[tex]Mode = 2[/tex]
Solving (d): The Range
[tex]Range = Highest\ Data - Lowest\ Data[/tex]
The highest is 9 and the lowest is 1,
So:
[tex]Range = 9 - 1[/tex]
[tex]Range = 8[/tex]
Solving (e): Lower (Q1) and Upper (Q3) quartile.
Q1 is calculated as:
[tex]Q_1 = \frac{1}{4}(n+1)th[/tex]
[tex]Q_1 = \frac{1}{4}(30+1)th[/tex]
[tex]Q_1 = \frac{1}{4}(31)th[/tex]
[tex]Q_1 = 7.75th[/tex]
[tex]Q_1 = 8th\ item[/tex]
[tex]Q_1 = 1[/tex] -- from the arranged data
Q3 is calculated as:
[tex]Q_3 = \frac{3}{4}(n+1)th[/tex]
[tex]Q_3 = \frac{3}{4}(30+1)th[/tex]
[tex]Q_3 = \frac{3}{4}(31)th[/tex]
[tex]Q_3 = 23.25th[/tex]
[tex]Q_3 = 23rd\ item[/tex]
[tex]Q_3 = 3[/tex] -- from the arranged data
Solving (f): The interquartile range (IQR)
[tex]IQR = Q_3 - Q_1[/tex]
[tex]IQR = 3-1[/tex]
[tex]IQR = 2[/tex]
