Q. 12 - Elementary Statistics

Question

A quantitative data set has been grouped by using cutpoint grouping with equal-width classes of width 8.

a)  If the midpoint of the first class is 10, what are its lower and upper cutpoints?

b) What is the class midpoint of the second class?

c) What are the lower and upper cutpoints of the third class?

d) Which class would contain an observation of 22? 

Step-by-Step Solution

Answer

Ans:  

(a)  The first class is $6-14$.

(b)  The midpoint of the second class is $18$.

(c)  The lower and upper cutpoints of the third class are $22 and 30$.

(d)  The third class contains observation $22$.

Step 1. Given information.

given,

    A quantitative data set has been grouped by using cutpoint grouping with equal-width classes of width 8. 

Step 2. A set of quantitative data is collected using data-grouping scores equal to 8.

a) If the middle area of the class is 10 then we get the cut and bottom points shown below:

By definition, the class midpoint is given by,

   $$\begin{gathered} Midpoint= \frac{Lower cutpoint + Upper cutpoint }{2} \\ \Rightarrow 10 =\frac{Lower cutpoint + Upper cutpoint }{2} \end{gathered}$$

$Lower cutpoint + Upper cutpoint =20 ...\left(1\right)$

Given that the width of the class interval is 8
 That is,

    $Upper cutpoint - Lower cutpoint =8 \dots \left(2\right)$

Step 3. Solving the equation (1) and (2), we get

   $$\begin{gathered} 2 Upper cutpoint =28 \\ Upper cutpoin =\frac{28}{2} =14 \end{gathered}$$

By substituting the Upper cutpoint in equation (1), we get

   $$\begin{gathered} Lower cutpoint + Upper cutpoint =20 \\ \Rightarrow Lower cutpoint =20 - Upper cutpoint \\ = 20-14 = 6 \end{gathered}$$

Therefore, the first class is $6-14$.

Step 4. (b) Now, find the second class midpoint shown below:

The lower cutpoint for the first class is $6$.

The upper cutpoint for the first class is $14$.

In Cutpoint integration the lower cut area of the second phase is equal to the uppercut area of the first phase.

So the Lower cutpoint for the second phase is $14$.

The top cutpoint for the second phase is,

    $$\begin{gathered} = Lower cutpoint of the second class + Class width \\ =14+8 =22 \end{gathered}$$

By definition, the midpoint is given by,

    $$\begin{gathered} Midpoint=\frac{Lower cutpoint + Upper cutpoint}{2} \\ =\frac{14+22}{2} =18 \end{gathered}$$

Therefore, the midpoint of the second class is 18.

Step 5. (c) We find the lower and cutpoints of the third class is shown below:

 The Lower cutpoint of the second class is $14$.

The Upper cutpoint of the second class is $22$.

In Cutpoint collection the lower cut area of the third phase is equal to the uppercut area of the second phase.

So the Lower cutpoint of the third class is $22$
 The upper cutpoint of the third Class is,

    $$\begin{gathered} = Lower cutpoint of the third class + Class width \\ =22+8 =30 \end{gathered}$$

Therefore, the lower and upper cutpoints of the third class are $22 and 30$.

Step 6. (d) Now,

The lower and upper cutpoints of the third class are $22 and 30$.

Thus, the third class contains observation $8$.