Find the best answers to your questions with the help of IDNLearn.com's knowledgeable users. Ask your questions and receive comprehensive and trustworthy answers from our experienced community of professionals.
Sagot :
To find the mean of the given frequency distribution, we follow these steps:
1. Determine the midpoints of each temperature interval:
Each temperature interval is given as:
[tex]\(40-44\)[/tex]
[tex]\(45-49\)[/tex]
[tex]\(50-54\)[/tex]
[tex]\(55-59\)[/tex]
* [tex]\(60-64\)[/tex]
The midpoints are calculated by averaging the endpoints of each interval. Therefore:
[tex]\[ \begin{align*} \text{Midpoint of } 40-44 &: \frac{40 + 44}{2} = 42 \\ \text{Midpoint of } 45-49 &: \frac{45 + 49}{2} = 47 \\ \text{Midpoint of } 50-54 &: \frac{50 + 54}{2} = 52 \\ \text{Midpoint of } 55-59 &: \frac{55 + 59}{2} = 57 \\ \text{Midpoint of } 60-64 &: \frac{60 + 64}{2} = 62 \\ \end{align*} \][/tex]
2. List the frequencies corresponding to each interval:
The given frequencies are:
[tex]\( f_1 = 1 \)[/tex] for [tex]\(40-44\)[/tex]
[tex]\( f_2 = 4 \)[/tex] for [tex]\(45-49\)[/tex]
[tex]\( f_3 = 9 \)[/tex] for [tex]\(50-54\)[/tex]
[tex]\( f_4 = 7 \)[/tex] for [tex]\(55-59\)[/tex]
* [tex]\( f_5 = 1 \)[/tex] for [tex]\(60-64\)[/tex]
3. Compute the product of each midpoint and its corresponding frequency:
[tex]\[ \begin{align*} 42 \times 1 &= 42 \\ 47 \times 4 &= 188 \\ 52 \times 9 &= 468 \\ 57 \times 7 &= 399 \\ 62 \times 1 &= 62 \\ \end{align*} \][/tex]
4. Sum the frequencies and sum the products of midpoints and frequencies:
The sum of frequencies ([tex]\(\Sigma f\)[/tex]) is:
[tex]\[ 1 + 4 + 9 + 7 + 1 = 22 \][/tex]
The sum of the products ([tex]\(\Sigma (f \cdot x)\)[/tex]) is:
[tex]\[ 42 + 188 + 468 + 399 + 62 = 1159 \][/tex]
5. Calculate the mean ([tex]\(\mu\)[/tex]) using the formula for the mean of grouped data:
[tex]\[ \text{Mean} = \frac{\Sigma (f \cdot x)}{\Sigma f} = \frac{1159}{22} \approx 52.68 \][/tex]
6. Round the mean to the nearest tenth:
[tex]\[ \text{Mean} \approx 52.7 \][/tex]
So, the mean of the frequency distribution is [tex]\( 52.7 \)[/tex] degrees.
Comparing this computed mean to the actual mean of 52.8 degrees, the values are very close, showing that the calculation is accurate.
Thus, the mean of the frequency distribution is
[tex]\[ \boxed{52.7} \text{ degrees (rounded to the nearest tenth).} \][/tex]
1. Determine the midpoints of each temperature interval:
Each temperature interval is given as:
[tex]\(40-44\)[/tex]
[tex]\(45-49\)[/tex]
[tex]\(50-54\)[/tex]
[tex]\(55-59\)[/tex]
* [tex]\(60-64\)[/tex]
The midpoints are calculated by averaging the endpoints of each interval. Therefore:
[tex]\[ \begin{align*} \text{Midpoint of } 40-44 &: \frac{40 + 44}{2} = 42 \\ \text{Midpoint of } 45-49 &: \frac{45 + 49}{2} = 47 \\ \text{Midpoint of } 50-54 &: \frac{50 + 54}{2} = 52 \\ \text{Midpoint of } 55-59 &: \frac{55 + 59}{2} = 57 \\ \text{Midpoint of } 60-64 &: \frac{60 + 64}{2} = 62 \\ \end{align*} \][/tex]
2. List the frequencies corresponding to each interval:
The given frequencies are:
[tex]\( f_1 = 1 \)[/tex] for [tex]\(40-44\)[/tex]
[tex]\( f_2 = 4 \)[/tex] for [tex]\(45-49\)[/tex]
[tex]\( f_3 = 9 \)[/tex] for [tex]\(50-54\)[/tex]
[tex]\( f_4 = 7 \)[/tex] for [tex]\(55-59\)[/tex]
* [tex]\( f_5 = 1 \)[/tex] for [tex]\(60-64\)[/tex]
3. Compute the product of each midpoint and its corresponding frequency:
[tex]\[ \begin{align*} 42 \times 1 &= 42 \\ 47 \times 4 &= 188 \\ 52 \times 9 &= 468 \\ 57 \times 7 &= 399 \\ 62 \times 1 &= 62 \\ \end{align*} \][/tex]
4. Sum the frequencies and sum the products of midpoints and frequencies:
The sum of frequencies ([tex]\(\Sigma f\)[/tex]) is:
[tex]\[ 1 + 4 + 9 + 7 + 1 = 22 \][/tex]
The sum of the products ([tex]\(\Sigma (f \cdot x)\)[/tex]) is:
[tex]\[ 42 + 188 + 468 + 399 + 62 = 1159 \][/tex]
5. Calculate the mean ([tex]\(\mu\)[/tex]) using the formula for the mean of grouped data:
[tex]\[ \text{Mean} = \frac{\Sigma (f \cdot x)}{\Sigma f} = \frac{1159}{22} \approx 52.68 \][/tex]
6. Round the mean to the nearest tenth:
[tex]\[ \text{Mean} \approx 52.7 \][/tex]
So, the mean of the frequency distribution is [tex]\( 52.7 \)[/tex] degrees.
Comparing this computed mean to the actual mean of 52.8 degrees, the values are very close, showing that the calculation is accurate.
Thus, the mean of the frequency distribution is
[tex]\[ \boxed{52.7} \text{ degrees (rounded to the nearest tenth).} \][/tex]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.