Question

Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 50.7 degrees. Low Temperature ​(circle​F) 40minus44 45minus49 50minus54 55minus59 60minus64 Frequency 2 7 9 5 1 The mean of the frequency distribution is nothing degrees. ​(Round to the nearest tenth as​ needed.)

Answer 1

Answer: The mean of the frequency distribution is 51.7°.

The computed mean 51.7° is greater than the actual mean 50.7° .

Step-by-step explanation:

The given table :

  Low Temperature ​(°​F)      Frequency     Mid -value  

      Class interval                  f                        x                  fx

          40-44                           2                     42.5               85

          45-49                           7                     47.5               95                      

          50-54                          9                     52.5               105

           55-59                         5                     57.5                115

           60-64                          1                      62.5               25

$\text{Now, Mean}=\dfrac{\sum fx}{\sum f}\\\\=\dfrac{85+332.5+472.5+287.5+62.5}{24}\\\\=\dfrac{1240}{24}=51.6666666667\approx51.7$

The computed mean 51.7° is greater than the actual mean 50.7° .

Answer 2

The mean of a distribution is the average of the distribution ; it is the sum of observation divided by the count of the observation.

• The mean of the frequency distribution is $51.2^oF$
• The computed mean $(51.2^oF)$ is greater than the actual mean $(50.7^oF)$

Given that:

$\begin{array}{cc}{Low\ Temperature(^oF) & Frequency&40-44&2\\45-49&7&50-54&9&55-59&5&60-64&1\end{array}$

First, we calculate the class midpoint.

This is the sum of the class interval divided by 2.

The midpoint of class 40 - 44 is $\frac{40+44}{2} = 42$

The midpoint of class 45 - 49 is $\frac{45+49}{2} = 47$

The midpoint of class 50 - 54 is $\frac{50+54}{2} = 52$

The midpoint of class 55 - 59 is $\frac{55+59}{2} = 57$

The midpoint of class 60 - 64 is $\frac{60+64}{2} = 62$

So, the table becomes:

$\begin{array}{ccc}{Low\ Temperature(^oF) & Frequency (f)&x&40-44&2&42\\45-49&7&47&50-54&9&52&55-59&5&57&60-64&1&62\end{array}$

The mean is then calculated as:

$\bar x = \frac{\sum fx}{\sum f}$

So, we have:

$\bar x = \frac{2 \times 42 + 7 \times 47 + 9 \times 52 + 5 \times 57 + 1 \times 62}{2+7+9+5+1}$

$\bar x = \frac{1228}{24}$

$\bar x = 51.2$

Hence, the mean of the frequency distribution is $51.2^oF$

By comparing the computed mean and the actual mean, we can conclude that the computed mean $(51.2^oF)$ is greater than the actual mean $(50.7^oF)$

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