Answer:
See explanation
Step-by-step explanation:
The question says the diameters of 50 pillars were measured. However, only the measurements of 20 pillars were provided.
So, I will work with 20
Given
[tex]n = 20[/tex]
[tex]62\ 74\ 92\ 86\ 92[/tex]
[tex]106\ 118\ 68\ 75\ 101[/tex]
[tex]93\ 98\ 69\ 80\ 87[/tex]
[tex]101\ 77\ 96\ 90\ 97[/tex]
Solving (a): Frequency distribution table
Using the given intervals:
[tex](60,70]\ => 61 - 70[/tex]
[tex](70, 80] =>71 - 80[/tex]
[tex](80,901]\ =>81 - 90[/tex]
[tex](90,100] =>91 - 100[/tex]
[tex](100, 110] =>101 - 110[/tex]
[tex](110,120] => 111 - 120[/tex]
The table is as follows:
[tex]\begin{array}{ccc}{Interval} & {Class\ Mark (x)} & {Frequency (f)} & {61-70} & {65.5} & {3} \ \\ {71-80} & {75.5} & {3} & {81-90} & {85.5} & {3} & {91-100} & {95.5} & {7} & {101-110} & {105.5} & {3} & {111-120} & {115.5} & {1} \ \end{array}[/tex]
[tex]Total \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 20[/tex]
The class mark is the average of the class interval
For instance: The class mark of 61 - 70 is:
[tex]x = \frac{1}{2}(61 + 70) = \frac{1}{2} * 131 = 65.5[/tex]
This is applied to other intervals
Solving (b): The histogram
See attachment for histogram
The frequency is plotted on the y-axis while the class interval, on the x-axis.
