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?

Short 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 $22and30$.

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

Step-by-step solution

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,

$$\begin{gathered} Midpoint=\frac{Lowercutpoint+Uppercutpoint}{2} \\ \Rightarrow 10=\frac{Lowercutpoint+Uppercutpoint}{2} \end{gathered}$$

$Lowercutpoint+Uppercutpoint=20...\left(1\right)$

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

$Uppercutpoint-Lowercutpoint=8\dots \left(2\right)$

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

$$\begin{gathered} 2Uppercutpoint=28 \\ Uppercutpoin=\frac{28}{2}=14 \end{gathered}$$

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

$$\begin{gathered} Lowercutpoint+Uppercutpoint=20 \\ \Rightarrow Lowercutpoint=20-Uppercutpoint \\ =20-14=6 \end{gathered}$$

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,

$$\begin{gathered} =Lowercutpointofthesecondclass+Classwidth \\ =14+8=22 \end{gathered}$$

By definition, the midpoint is given by,

$$\begin{gathered} Midpoint=\frac{Lowercutpoint+Uppercutpoint}{2} \\ =\frac{14+22}{2}=18 \end{gathered}$$

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,

$$\begin{gathered} =Lowercutpointofthethirdclass+Classwidth \\ =22+8=30 \end{gathered}$$

Therefore, the lower and upper cutpoints of the third class are$22and30$.

Step 6. (d)  Now,

The lower and upper cutpoints of the third class are$22and30$.

Thus, the third class contains observation $8$.