Questions: Often, frequency distributions are reported using unequal class widths because the frequencies of some groups would otherwise be small or very large. Consider the following data, which represent the daytime household temperature the thermostat is set to when someone is home for a random sample of 754 households. Determine the class midpoint, if necessary, for each class and approximate the mean and standard deviation temperature. Class Class Midpoint 61-64 65-67 68-69 70 71-72 73-76 77-80 (Round to one decimal place as needed.) The sample mean is °F. (Round to one decimal place as needed.) The sample standard deviation is °F. (Round to one decimal place as needed.)

Often, frequency distributions are reported using unequal class widths because the frequencies of some groups would otherwise be small or very large. Consider the following data, which represent the daytime household temperature the thermostat is set to when someone is home for a random sample of 754 households. Determine the class midpoint, if necessary, for each class and approximate the mean and standard deviation temperature.

Class Class Midpoint
61-64
65-67
68-69
70
71-72
73-76
77-80

(Round to one decimal place as needed.)
The sample mean is °F.
(Round to one decimal place as needed.)
The sample standard deviation is °F.
(Round to one decimal place as needed.)

Transcript text: Often, frequency distributions are reported using unequal class widths because the frequencies of some groups would otherwise be small or very large. Consider the following data, which represent the daytime household temperature the thermostat is set to when someone is home for a random sample of 754 households. Determine the class midpoint, if necessary, for each class and approximate the mean and standard deviation temperature. \begin{tabular}{cc} Class & Class Midpoint \\ \hline $61-64$ & $\square$ \\ \hline $65-67$ & $\square$ \\ \hline $68-69$ & $\square$ \\ \hline 70 & $\square$ \\ \hline $71-72$ & $\square$ \\ \hline $73-76$ & $\square$ \\ \hline $77-80$ & $\square$ \\ \hline \end{tabular} (Round to one decimal place as needed.) The sample mean is $\square$ ${ }^{\circ} \mathrm{F}$. (Round to one decimal place as needed.) The sample standard deviation is $\square$ ${ }^{\circ} \mathrm{F}$. (Round to one decimal place as needed.)

Solution

Solution Steps

Step 1: Calculate Class Midpoints

For each temperature class, the midpoint is calculated. If the class is a range, the midpoint is the average of the lower and upper bounds. If the class represents a single value, the midpoint is that value itself. Class Midpoints: [62.5, 66, 68.5, 70, 71.5, 74.5, 78.5]

Step 2: Compute Weighted Mean

The weighted mean of the temperatures is calculated using the class midpoints and their corresponding frequencies. Weighted Mean: $\bar{x} = \frac{\sum{f_i x_i}}{N} = 0 $, where $f_i$ is the frequency of the $i^{th}$ class, $x_i$ is the midpoint of the $i^{th}$ class, and $N$ is the total sample size.

Step 3: Calculate Standard Deviation

The standard deviation is approximated using the formula for the standard deviation of a frequency distribution. Standard Deviation: $\sigma = \sqrt{\frac{\sum{f_i (x_i - \bar{x})^2}}{N}} = 0 $, where $f_i$ is the frequency of the $i^{th}$ class, $x_i$ is the midpoint of the $i^{th}$ class, and $\bar{x}$ is the weighted mean.