Questions: An energy study found that household use of electricity had a mean of 682 and a standard deviation of 551 kilowatt-hours. Complete parts (a) and (b). a. Suppose the distribution of energy use was normal. Using a table, calculator, or software that can give normal probabilities, find the proportion of households with electricity use greater than 1000 kilowatt-hours. The proportion of households with electricity use greater than 1000 kilowatt-hours is 0.28 . (Round to the nearest hundredth as needed.) b. Based on the mean and standard deviation given, do you think that the distribution of energy use actually is normal? Why or why not? No, the distribution of energy use does not appear to be normal since 0 has a z-score of only (0-682) / 551=-1.24, yet energy use cannot be negative Yes, the distribution of energy use appears to be normal since the value 682 - 551=131 is positive.

An energy study found that household use of electricity had a mean of 682 and a standard deviation of 551 kilowatt-hours. Complete parts (a) and (b).
a. Suppose the distribution of energy use was normal. Using a table, calculator, or software that can give normal probabilities, find the proportion of households with electricity use greater than 1000 kilowatt-hours.

The proportion of households with electricity use greater than 1000 kilowatt-hours is 0.28 . (Round to the nearest hundredth as needed.)
b. Based on the mean and standard deviation given, do you think that the distribution of energy use actually is normal? Why or why not?
No, the distribution of energy use does not appear to be normal since 0 has a z-score of only (0-682) / 551=-1.24, yet energy use cannot be negative
Yes, the distribution of energy use appears to be normal since the value 682 - 551=131 is positive.

Transcript text: An energy study found that household use of electricity had a mean of 682 and a standard deviation of 551 kilowatt-hours. Complete parts (a) and (b). a. Suppose the distribution of energy use was normal. Using a table, calculator, or software that can give normal probabilities, find the proportion of households with electricity use greater than 1000 kilowatt-hours. The proportion of households with electricity use greater than 1000 kilowatt-hours is 0.28 . (Round to the nearest hundredth as needed.) b. Based on the mean and standard deviation given, do you think that the distribution of energy use actually is normal? Why or why not? No, the distribution of energy use does not appear to be normal since 0 has a 2 -score of only $(0-682) / 551=-1.24$, yet energy use cannot be negative Yes, the distribution of energy use appears to be normal since the value 682 - $551=131$ is positive.

Solution

Solution Steps

Step 1: Calculate the Probability of Energy Use Greater than 1000 kWh

To find the proportion of households with electricity use greater than 1000 kWh, we first calculate the cumulative probability of energy use being less than or equal to 1000 kWh. This is given by:

\[ P(X \leq 1000) = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(0.5771) - \Phi(-\infty) = 0.7181 \]

Thus, the probability of energy use being greater than 1000 kWh is:

\[ P(X > 1000) = 1 - P(X \leq 1000) = 1 - 0.7181 = 0.2819 \]

Step 2: Calculate the Z-Score for 0 kWh

Next, we calculate the z-score for 0 kWh using the formula:

\[ z = \frac{X - \mu}{\sigma} = \frac{0 - 682}{551} = -1.2377 \]

Step 3: Assess the Normality of the Distribution

Based on the calculated z-score of $ z = -1.2377 $, we analyze whether the distribution of energy use can be considered normal. Since the z-score for 0 kWh is not extremely low (e.g., less than -3), we conclude that:

Yes, the distribution of energy use appears to be normal since the value $ 682 - 551 = 131 $ is positive.

Final Answer

The proportion of households with energy use greater than 1000 kWh is approximately $ 0.2819 $, and the distribution of energy use appears to be normal.

\[ \boxed{P(X > 1000) \approx 0.2819} \]

\[ \text{Distribution normality: Yes} \]

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