Answer :
Sure, let's solve this step by step.
1. Understand the given data:
- We have the electric usage data of 200 homes.
- The data is summarized in a frequency table which pairs kilowatt-hours (kWh) with their respective frequencies (number of homes).
[tex]\[ \begin{array}{|c|c|} \hline \text{kWh} & \text{Frequency} \\ \hline 500 & 7 \\ \hline 600 & 20 \\ \hline 700 & 26 \\ \hline 1000 & 50 \\ \hline 1100 & 34 \\ \hline 1200 & 12 \\ \hline \end{array} \][/tex]
- The standard deviation of the data is given as 156.9323.
2. Extract the relevant numbers from the table:
- Kilowatt-hours: [tex]\( 500, 600, 700, 1000, 1100, 1200 \)[/tex]
- Frequencies: [tex]\( 7, 20, 26, 50, 34, 12 \)[/tex]
3. Calculate the mean ([tex]\(\bar{x}\)[/tex]):
- The mean is calculated using the formula:
[tex]\[ \bar{x} = \frac{\sum{(x_i \cdot f_i)}}{\sum{f_i}} \][/tex]
where [tex]\(x_i\)[/tex] are the kilowatt-hours and [tex]\(f_i\)[/tex] are their respective frequencies.
[tex]\[ \bar{x} = \frac{(500 \cdot 7) + (600 \cdot 20) + (700 \cdot 26) + (1000 \cdot 50) + (1100 \cdot 34) + (1200 \cdot 12)}{7 + 20 + 26 + 50 + 34 + 12} \][/tex]
Calculating the numerator (sum of the products of kWh and frequencies):
[tex]\[ = 3500 + 12000 + 18200 + 50000 + 37400 + 14400 = 135500 \][/tex]
Calculating the denominator (sum of frequencies):
[tex]\[ = 7 + 20 + 26 + 50 + 34 + 12 = 149 \][/tex]
Therefore, the mean usage is:
[tex]\[ \bar{x} = \frac{135500}{149} \approx 909.3959731543624 \, \text{kWh} \][/tex]
4. Calculate the variance ([tex]\(\sigma^2\)[/tex]):
- Variance is calculated using the formula:
[tex]\[ \sigma^2 = \frac{\sum{f_i (x_i - \bar{x})^2}}{\sum{f_i}} \][/tex]
where [tex]\(x_i\)[/tex] are the kilowatt-hours, [tex]\(f_i\)[/tex] are the frequencies, and [tex]\(\bar{x}\)[/tex] is the mean.
[tex]\[ \sigma^2 = \frac{7 \cdot (500 - 909.3959731543624)^2 + 20 \cdot (600 - 909.3959731543624)^2 + 26 \cdot (700 - 909.3959731543624)^2 + 50 \cdot (1000 - 909.3959731543624)^2 + 34 \cdot (1100 - 909.3959731543624)^2 + 12 \cdot (1200 - 909.3959731543624)^2}{149} \][/tex]
Simplifying this:
[tex]\[ \sigma^2 = \frac{(7 \cdot (409.3959731543624)^2) + (20 \cdot (309.3959731543624)^2) + (26 \cdot (209.3959731543624)^2) + (50 \cdot (90.6040268456376)^2) + (34 \cdot (190.6040268456376)^2) + (12 \cdot (290.6040268456376)^2)}{149} \][/tex]
After calculating each term and summing them up, we find the sum to be approximately:
[tex]\[ \approx 6886861.63 \][/tex]
Therefore, the variance is:
[tex]\[ \sigma^2 = \frac{6886861.63}{149} \approx 46220.440520697266 \, \text{kWh}^2 \][/tex]
5. Reconfirm the standard deviation:
- The standard deviation ([tex]\(\sigma\)[/tex]) is the square root of the variance.
[tex]\[ \sigma \approx \sqrt{46220.440520697266} \approx 156.9323 \, \text{kWh} \][/tex]
Thus, the mean is approximately 909.396 kWh, the variance is approximately 46220.441 kWh², and the standard deviation is approximately 156.9323 kWh.
1. Understand the given data:
- We have the electric usage data of 200 homes.
- The data is summarized in a frequency table which pairs kilowatt-hours (kWh) with their respective frequencies (number of homes).
[tex]\[ \begin{array}{|c|c|} \hline \text{kWh} & \text{Frequency} \\ \hline 500 & 7 \\ \hline 600 & 20 \\ \hline 700 & 26 \\ \hline 1000 & 50 \\ \hline 1100 & 34 \\ \hline 1200 & 12 \\ \hline \end{array} \][/tex]
- The standard deviation of the data is given as 156.9323.
2. Extract the relevant numbers from the table:
- Kilowatt-hours: [tex]\( 500, 600, 700, 1000, 1100, 1200 \)[/tex]
- Frequencies: [tex]\( 7, 20, 26, 50, 34, 12 \)[/tex]
3. Calculate the mean ([tex]\(\bar{x}\)[/tex]):
- The mean is calculated using the formula:
[tex]\[ \bar{x} = \frac{\sum{(x_i \cdot f_i)}}{\sum{f_i}} \][/tex]
where [tex]\(x_i\)[/tex] are the kilowatt-hours and [tex]\(f_i\)[/tex] are their respective frequencies.
[tex]\[ \bar{x} = \frac{(500 \cdot 7) + (600 \cdot 20) + (700 \cdot 26) + (1000 \cdot 50) + (1100 \cdot 34) + (1200 \cdot 12)}{7 + 20 + 26 + 50 + 34 + 12} \][/tex]
Calculating the numerator (sum of the products of kWh and frequencies):
[tex]\[ = 3500 + 12000 + 18200 + 50000 + 37400 + 14400 = 135500 \][/tex]
Calculating the denominator (sum of frequencies):
[tex]\[ = 7 + 20 + 26 + 50 + 34 + 12 = 149 \][/tex]
Therefore, the mean usage is:
[tex]\[ \bar{x} = \frac{135500}{149} \approx 909.3959731543624 \, \text{kWh} \][/tex]
4. Calculate the variance ([tex]\(\sigma^2\)[/tex]):
- Variance is calculated using the formula:
[tex]\[ \sigma^2 = \frac{\sum{f_i (x_i - \bar{x})^2}}{\sum{f_i}} \][/tex]
where [tex]\(x_i\)[/tex] are the kilowatt-hours, [tex]\(f_i\)[/tex] are the frequencies, and [tex]\(\bar{x}\)[/tex] is the mean.
[tex]\[ \sigma^2 = \frac{7 \cdot (500 - 909.3959731543624)^2 + 20 \cdot (600 - 909.3959731543624)^2 + 26 \cdot (700 - 909.3959731543624)^2 + 50 \cdot (1000 - 909.3959731543624)^2 + 34 \cdot (1100 - 909.3959731543624)^2 + 12 \cdot (1200 - 909.3959731543624)^2}{149} \][/tex]
Simplifying this:
[tex]\[ \sigma^2 = \frac{(7 \cdot (409.3959731543624)^2) + (20 \cdot (309.3959731543624)^2) + (26 \cdot (209.3959731543624)^2) + (50 \cdot (90.6040268456376)^2) + (34 \cdot (190.6040268456376)^2) + (12 \cdot (290.6040268456376)^2)}{149} \][/tex]
After calculating each term and summing them up, we find the sum to be approximately:
[tex]\[ \approx 6886861.63 \][/tex]
Therefore, the variance is:
[tex]\[ \sigma^2 = \frac{6886861.63}{149} \approx 46220.440520697266 \, \text{kWh}^2 \][/tex]
5. Reconfirm the standard deviation:
- The standard deviation ([tex]\(\sigma\)[/tex]) is the square root of the variance.
[tex]\[ \sigma \approx \sqrt{46220.440520697266} \approx 156.9323 \, \text{kWh} \][/tex]
Thus, the mean is approximately 909.396 kWh, the variance is approximately 46220.441 kWh², and the standard deviation is approximately 156.9323 kWh.
