Question

Below is a frequency distribution for our RCCC sample of men's heights.

Class Class Frequency Midpoint (inches)
64-65 4
66-67 4
68-69 4
70-71 5
72-73 10
74-75 4
a. Fill in the midpoint of each class in the column provided.
b. Enter the midpoints in L, and the frequencies in L2, and use 1- VarStats to calculate the mean and standard deviation of the frequency distribution. Using the frequency distribution, I found the mean height to be ___________with a standard deviation of____________
c. Now, let's compare the mean of the frequency distribution you just found in part (b), which is an estimate of the actual sample mean, to the actual sample mean you found in (#1).
Using the frequency distribution in (a), I found the mean height to be___________ inches, while using the actual data in #1, I found the mean height to be___________ inches. The true sample mean (using all the data) and the sample mean estimated from a frequency distribution can be fairly close to each other or very different. In this case, do you think the two means were close___________ Using complete sentences, explain why you think the two means came out so close, or why they came out so different, whichever the case may be.

Answer 1

`\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & { 64.5} & {64-65} & {4} & {66.5 } & {66-67} & {4} & {68.5 } & {68-69} & {4} &{70.5 } & {70-71} & {5} & {72.5 } & {72-73} & {10} & {74.5 } & {74-75} & {4}\ \end{array}`

Using the frequency distribution, I found the mean height to be 70.1129 with a standard deviation of 3.2831

Explanation:

Given

`\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & { } & {64-65} & {4} & { } & {66-67} & {4} & { } & {68-69} & {4} &{ } & {70-71} & {5} & { } & {72-73} & {10} & { } & {74-75} & {4}\ \end{array}`

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

`M = (1)/(2)(Lower + Upper)`

Where

`Lower \to` Lower class interval

`Upper \to` Upper class interval

So, we have:

Class 64-65:

`M = (1)/(2)(64 + 65) = 64.5`

Class 66 - 67:

`M = (1)/(2)(66 + 67) = 66.5`

When the computation is completed, the frequency distribution will be:

`\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & { 64.5} & {64-65} & {4} & {66.5 } & {66-67} & {4} & {68.5 } & {68-69} & {4} &{70.5 } & {70-71} & {5} & {72.5 } & {72-73} & {10} & {74.5 } & {74-75} & {4}\ \end{array}`

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

`\bar x = 70.1129`

`\sigma = 3.2831`

See attachment for result of 1-VarStats

Solving (c): Compare the calculated mean to the actual mean

The actual mean is missing from the question, so I will make assumptions in this part

Assume they are close

This means that the selected sample is a reflection of the actual population where the samples are selected.

Assume they are not close

This means that the selected sample does not reflect the actual population where the samples are selected.