Mathematics, 31.07.2019 20:20 nengliangli523

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 (circlef) 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.)

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Answer from: adjjones2011

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)

Read more about mean of the frequency distribution at:

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Answer from: amaljiiju5022

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 from: Quest

a negative exponent is translated as follows:

$a^{-m} = \dfrac{1}{a^m}$

so, we have

$\left(5^{-1}+3^{-1}+2^{-1}\right)^{-1} = \left(\dfrac{1}{5}+\dfrac{1}{3}+\dfrac{1}{2}\right)^{-1} = \left(\dfrac{6+10+15}{30}\right)^{-1}=\left(\dfrac{31}{30}\right)^{-1}=\dfrac{30}{31}$

Answer from: Quest

d. vertex

step-by-step explanation:

Multiple choice: select the best answer and click "submit." what is the name of the sides of the t

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Find the mean of the data summarized in the given frequency distribution. compare the computed mean...