Answer :
Let's break down the steps to find the mean, variance, and standard deviation for the given frequency distribution.
The class intervals and their corresponding frequencies are:
- [tex]\(0-9\)[/tex] with frequency [tex]\(15\)[/tex]
- [tex]\(10-19\)[/tex] with frequency [tex]\(15\)[/tex]
- [tex]\(20-29\)[/tex] with frequency [tex]\(23\)[/tex]
- [tex]\(30-39\)[/tex] with frequency [tex]\(22\)[/tex]
- [tex]\(40-49\)[/tex] with frequency [tex]\(25\)[/tex]
- [tex]\(50-59\)[/tex] with frequency [tex]\(10\)[/tex]
- [tex]\(60-69\)[/tex] with frequency [tex]\(5\)[/tex]
- [tex]\(70-79\)[/tex] with frequency [tex]\(10\)[/tex]
Step 1: Determining Midpoints
First, calculate the midpoints ([tex]\(x_i\)[/tex]) of each class interval:
- Midpoint of [tex]\(0-9\)[/tex] is [tex]\(\frac{0+9}{2} = 4.5\)[/tex]
- Midpoint of [tex]\(10-19\)[/tex] is [tex]\(\frac{10+19}{2} = 14.5\)[/tex]
- Midpoint of [tex]\(20-29\)[/tex] is [tex]\(\frac{20+29}{2} = 24.5\)[/tex]
- Midpoint of [tex]\(30-39\)[/tex] is [tex]\(\frac{30+39}{2} = 34.5\)[/tex]
- Midpoint of [tex]\(40-49\)[/tex] is [tex]\(\frac{40+49}{2} = 44.5\)[/tex]
- Midpoint of [tex]\(50-59\)[/tex] is [tex]\(\frac{50+59}{2} = 54.5\)[/tex]
- Midpoint of [tex]\(60-69\)[/tex] is [tex]\(\frac{60+69}{2} = 64.5\)[/tex]
- Midpoint of [tex]\(70-79\)[/tex] is [tex]\(\frac{70+79}{2} = 74.5\)[/tex]
So the midpoints are: [tex]\([4.5, 14.5, 24.5, 34.5, 44.5, 54.5, 64.5, 74.5]\)[/tex].
Step 2: Calculating the Mean
The mean ([tex]\(\bar{x}\)[/tex]) of the distribution is given by:
[tex]\[ \text{Mean} = \frac{\sum (f_i \cdot x_i)}{\sum f_i} \][/tex]
where [tex]\(f_i\)[/tex] is the frequency of the [tex]\(i\)[/tex]-th class interval, and [tex]\(x_i\)[/tex] is the midpoint of the [tex]\(i\)[/tex]-th class interval.
The total frequency [tex]\(\sum f_i = 15 + 15 + 23 + 22 + 25 + 10 + 5 + 10 = 125\)[/tex].
The sum of the product of frequencies and midpoints:
[tex]\[ \sum (f_i \cdot x_i) = (15 \cdot 4.5) + (15 \cdot 14.5) + (23 \cdot 24.5) + (22 \cdot 34.5) + (25 \cdot 44.5) + (10 \cdot 54.5) + (5 \cdot 64.5) + (10 \cdot 74.5) \][/tex]
[tex]\[ = 67.5 + 217.5 + 563.5 + 759 + 1112.5 + 545 + 322.5 + 745 \][/tex]
[tex]\[ = 4332 \][/tex]
Thus, the mean is:
[tex]\[ \text{Mean} = \frac{4332}{125} = 34.66 \][/tex]
Step 3: Calculating the Variance
The variance ([tex]\(\sigma^2\)[/tex]) is given by:
[tex]\[ \text{Variance} = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i} \][/tex]
First, calculate [tex]\((x_i - \bar{x})^2\)[/tex]:
[tex]\[ (4.5 - 34.66)^2 = ( -30.16)^2 = 909.61 \][/tex]
[tex]\[ (14.5 - 34.66)^2 = ( -20.16)^2 = 406.41 \][/tex]
[tex]\[ (24.5 - 34.66)^2 = ( -10.16)^2 = 103.22 \][/tex]
[tex]\[ (34.5 - 34.66)^2 = ( -0.16)^2 = 0.03 \][/tex]
[tex]\[ (44.5 - 34.66)^2 = (9.84)^2 = 96.86 \][/tex]
[tex]\[ (54.5 - 34.66)^2 = (19.84)^2 = 393.42 \][/tex]
[tex]\[ (64.5 - 34.66)^2 = (29.84)^2 = 890.61 \][/tex]
[tex]\[ (74.5 - 34.66)^2 = (39.84)^2 = 1587.06 \][/tex]
Now, sum the products of these values and the respective frequencies:
[tex]\[ \sum (f_i (x_i - \bar{x})^2) = (15 \cdot 909.61) + (15 \cdot 406.41) + (23 \cdot 103.22) + (22 \cdot 0.03) + (25 \cdot 96.86) + (10 \cdot 393.42) + (5 \cdot 890.61) + (10 \cdot 1587.06) \][/tex]
[tex]\[ = 13644.15 + 6096.15 + 2374.06 + 0.66 + 2421.5 + 3934.2 + 4453.05 + 15870.6 \][/tex]
[tex]\[ = 39094.37 \][/tex]
Thus, the variance is:
[tex]\[ \text{Variance} = \frac{39094.37}{125} = 390.37\][/tex]
Step 4: Calculating the Standard Deviation
The standard deviation ([tex]\(\sigma\)[/tex]) is the square root of the variance:
[tex]\[ \text{Standard Deviation} = \sqrt{390.37} = 19.76 \][/tex]
Summary:
- The mean of the distribution is [tex]\(34.66\)[/tex]
- The variance of the distribution is [tex]\(390.37\)[/tex]
- The standard deviation of the distribution is [tex]\(19.76\)[/tex]
These are the final results for the given frequency distribution.
The class intervals and their corresponding frequencies are:
- [tex]\(0-9\)[/tex] with frequency [tex]\(15\)[/tex]
- [tex]\(10-19\)[/tex] with frequency [tex]\(15\)[/tex]
- [tex]\(20-29\)[/tex] with frequency [tex]\(23\)[/tex]
- [tex]\(30-39\)[/tex] with frequency [tex]\(22\)[/tex]
- [tex]\(40-49\)[/tex] with frequency [tex]\(25\)[/tex]
- [tex]\(50-59\)[/tex] with frequency [tex]\(10\)[/tex]
- [tex]\(60-69\)[/tex] with frequency [tex]\(5\)[/tex]
- [tex]\(70-79\)[/tex] with frequency [tex]\(10\)[/tex]
Step 1: Determining Midpoints
First, calculate the midpoints ([tex]\(x_i\)[/tex]) of each class interval:
- Midpoint of [tex]\(0-9\)[/tex] is [tex]\(\frac{0+9}{2} = 4.5\)[/tex]
- Midpoint of [tex]\(10-19\)[/tex] is [tex]\(\frac{10+19}{2} = 14.5\)[/tex]
- Midpoint of [tex]\(20-29\)[/tex] is [tex]\(\frac{20+29}{2} = 24.5\)[/tex]
- Midpoint of [tex]\(30-39\)[/tex] is [tex]\(\frac{30+39}{2} = 34.5\)[/tex]
- Midpoint of [tex]\(40-49\)[/tex] is [tex]\(\frac{40+49}{2} = 44.5\)[/tex]
- Midpoint of [tex]\(50-59\)[/tex] is [tex]\(\frac{50+59}{2} = 54.5\)[/tex]
- Midpoint of [tex]\(60-69\)[/tex] is [tex]\(\frac{60+69}{2} = 64.5\)[/tex]
- Midpoint of [tex]\(70-79\)[/tex] is [tex]\(\frac{70+79}{2} = 74.5\)[/tex]
So the midpoints are: [tex]\([4.5, 14.5, 24.5, 34.5, 44.5, 54.5, 64.5, 74.5]\)[/tex].
Step 2: Calculating the Mean
The mean ([tex]\(\bar{x}\)[/tex]) of the distribution is given by:
[tex]\[ \text{Mean} = \frac{\sum (f_i \cdot x_i)}{\sum f_i} \][/tex]
where [tex]\(f_i\)[/tex] is the frequency of the [tex]\(i\)[/tex]-th class interval, and [tex]\(x_i\)[/tex] is the midpoint of the [tex]\(i\)[/tex]-th class interval.
The total frequency [tex]\(\sum f_i = 15 + 15 + 23 + 22 + 25 + 10 + 5 + 10 = 125\)[/tex].
The sum of the product of frequencies and midpoints:
[tex]\[ \sum (f_i \cdot x_i) = (15 \cdot 4.5) + (15 \cdot 14.5) + (23 \cdot 24.5) + (22 \cdot 34.5) + (25 \cdot 44.5) + (10 \cdot 54.5) + (5 \cdot 64.5) + (10 \cdot 74.5) \][/tex]
[tex]\[ = 67.5 + 217.5 + 563.5 + 759 + 1112.5 + 545 + 322.5 + 745 \][/tex]
[tex]\[ = 4332 \][/tex]
Thus, the mean is:
[tex]\[ \text{Mean} = \frac{4332}{125} = 34.66 \][/tex]
Step 3: Calculating the Variance
The variance ([tex]\(\sigma^2\)[/tex]) is given by:
[tex]\[ \text{Variance} = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i} \][/tex]
First, calculate [tex]\((x_i - \bar{x})^2\)[/tex]:
[tex]\[ (4.5 - 34.66)^2 = ( -30.16)^2 = 909.61 \][/tex]
[tex]\[ (14.5 - 34.66)^2 = ( -20.16)^2 = 406.41 \][/tex]
[tex]\[ (24.5 - 34.66)^2 = ( -10.16)^2 = 103.22 \][/tex]
[tex]\[ (34.5 - 34.66)^2 = ( -0.16)^2 = 0.03 \][/tex]
[tex]\[ (44.5 - 34.66)^2 = (9.84)^2 = 96.86 \][/tex]
[tex]\[ (54.5 - 34.66)^2 = (19.84)^2 = 393.42 \][/tex]
[tex]\[ (64.5 - 34.66)^2 = (29.84)^2 = 890.61 \][/tex]
[tex]\[ (74.5 - 34.66)^2 = (39.84)^2 = 1587.06 \][/tex]
Now, sum the products of these values and the respective frequencies:
[tex]\[ \sum (f_i (x_i - \bar{x})^2) = (15 \cdot 909.61) + (15 \cdot 406.41) + (23 \cdot 103.22) + (22 \cdot 0.03) + (25 \cdot 96.86) + (10 \cdot 393.42) + (5 \cdot 890.61) + (10 \cdot 1587.06) \][/tex]
[tex]\[ = 13644.15 + 6096.15 + 2374.06 + 0.66 + 2421.5 + 3934.2 + 4453.05 + 15870.6 \][/tex]
[tex]\[ = 39094.37 \][/tex]
Thus, the variance is:
[tex]\[ \text{Variance} = \frac{39094.37}{125} = 390.37\][/tex]
Step 4: Calculating the Standard Deviation
The standard deviation ([tex]\(\sigma\)[/tex]) is the square root of the variance:
[tex]\[ \text{Standard Deviation} = \sqrt{390.37} = 19.76 \][/tex]
Summary:
- The mean of the distribution is [tex]\(34.66\)[/tex]
- The variance of the distribution is [tex]\(390.37\)[/tex]
- The standard deviation of the distribution is [tex]\(19.76\)[/tex]
These are the final results for the given frequency distribution.
