Respuesta :
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
[tex]\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[/tex]
The computed mean 51.7° is greater than the actual mean 50.7° .
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 [tex]51.2^oF[/tex]
- The computed mean [tex](51.2^oF)[/tex] is greater than the actual mean [tex](50.7^oF)[/tex]
Given that:
[tex]\begin{array}{cc}{Low\ Temperature(^oF) & Frequency&40-44&2\\45-49&7&50-54&9&55-59&5&60-64&1\end{array}[/tex]
First, we calculate the class midpoint.
This is the sum of the class interval divided by 2.
The midpoint of class 40 - 44 is [tex]\frac{40+44}{2} = 42[/tex]
The midpoint of class 45 - 49 is [tex]\frac{45+49}{2} = 47[/tex]
The midpoint of class 50 - 54 is [tex]\frac{50+54}{2} = 52[/tex]
The midpoint of class 55 - 59 is [tex]\frac{55+59}{2} = 57[/tex]
The midpoint of class 60 - 64 is [tex]\frac{60+64}{2} = 62[/tex]
So, the table becomes:
[tex]\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}[/tex]
The mean is then calculated as:
[tex]\bar x = \frac{\sum fx}{\sum f}[/tex]
So, we have:
[tex]\bar x = \frac{2 \times 42 + 7 \times 47 + 9 \times 52 + 5 \times 57 + 1 \times 62}{2+7+9+5+1}[/tex]
[tex]\bar x = \frac{1228}{24}[/tex]
[tex]\bar x = 51.2[/tex]
Hence, the mean of the frequency distribution is [tex]51.2^oF[/tex]
By comparing the computed mean and the actual mean, we can conclude that the computed mean [tex](51.2^oF)[/tex] is greater than the actual mean [tex](50.7^oF)[/tex]
Read more about mean of the frequency distribution at:
https://brainly.com/question/12269435