Solution
Since the outlier that is 26 has been removed
We will work with the remaining
Where X denotes the number of hours, and f represent the frequency corresponding to eaxh hours
We find the mean
The mean (X bar) is given by
[tex]\begin{gathered} mean=\frac{\Sigma fx}{\Sigma f} \\ mean=\frac{1(2)+2(2)+3(3)+4(3)+5(2)+6(2)}{2+2+3+3+2+2} \\ mean=\frac{2+4+9+12+10+12}{2+2+3+3+2+2} \\ mean=\frac{49}{14} \\ mean=\frac{7}{2} \\ mean=3.5 \end{gathered}[/tex]
We now find the median
Median is the middle number
Since the total frequency is 14
The median will be on the 7th and 8th term in ascending order
[tex]\begin{gathered} median=\frac{7th+8th}{2} \\ median=\frac{3+4}{2} \\ median=\frac{7}{2} \\ median=3.5 \end{gathered}[/tex]
Lastly, we will find the interquartile range
The formula is given by
[tex]IQR=Q_3-Q_1[/tex]
Where
[tex]\begin{gathered} Q_3=\frac{3}{4}(n+1)th\text{ term} \\ Q_1=\frac{1}{4}(n+1)th\text{ term} \end{gathered}[/tex]
We calculate for Q1 and Q3
[tex]\begin{gathered} Q_1=\frac{1}{4}(n+1)th\text{ term} \\ \text{n is the total frequency} \\ n=14 \\ Q_1=\frac{1}{4}(14+1)th\text{ term} \\ Q_1=\frac{1}{4}(15)th\text{ term} \\ Q_1=3.75th\text{ term} \\ Q_1\text{ falls betwe}en\text{ the frequency 3 and 4 in ascending order} \\ \text{From the table above} \\ Q_1=2 \end{gathered}[/tex][tex]\begin{gathered} Q_3=\frac{3}{4}(n+1)th\text{ term} \\ Q_3=\frac{3}{4}(14+1)th\text{ term} \\ Q_3=\frac{3}{4}(15)th\text{ term} \\ Q_3=11.25th\text{ term} \\ \text{From the table above} \\ Q_3=5 \end{gathered}[/tex]
Therefore, the IQR is
[tex]\begin{gathered} IQR=Q_3-Q_1 \\ IQR=5-2 \\ IQR=3 \end{gathered}[/tex]
