Question

question 9
1 pts
given the following information: \\( \mu = 50, \sigma = 12, n = 36 \\)
calculate \\( p(\bar{x} > 50.8)
.5279
.6554
.9918
.4721
.0082
.3446
question 10
1 pts
in order to estimate the average electric usage per month, a sample of 100 houses was selected, and the electric usage was determined. assume a population standard deviation of 480 - kilowatt hours. if the sample mean is 1,883 kwh, the 95% confidence interval estimate of the population mean is
(1787.768, 1978.232)
(1788.92, 1977.08)
(1804.04, 1961.96)
(1837.4, 1928.6)
(1803.32, 1962.68)
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Explanation:

Question 9

Step1: Find the standard error of the mean

The formula for the standard error of the mean ($\sigma_{\bar{x}}$) is $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$.
Given $\sigma = 12$ and $n = 36$, we have:
$\sigma_{\bar{x}} = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2$

Step2: Calculate the z-score

The z-score for the sample mean $\bar{x}$ is given by $z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}$.
We want to find $P(\bar{x} > 50.8)$, so $\bar{x} = 50.8$, $\mu = 50$, and $\sigma_{\bar{x}} = 2$.
$z = \frac{50.8 - 50}{2} = \frac{0.8}{2} = 0.4$

Step3: Find the probability using the z-table

We need $P(\bar{x} > 50.8)$, which is equivalent to $P(z > 0.4)$.
From the standard normal table, $P(z \leq 0.4) = 0.6554$.
Thus, $P(z > 0.4) = 1 - P(z \leq 0.4) = 1 - 0.6554 = 0.3446$? Wait, no—wait, let's check again. Wait, maybe I made a mistake. Wait, no, wait: Wait, the z-score is 0.4? Wait, 50.8 - 50 is 0.8, divided by 2 is 0.4. Then $P(z > 0.4)$ is 1 - 0.6554 = 0.3446? But the options include 0.3446 as an option. Wait, but let's recheck. Wait, maybe I messed up the direction. Wait, no, the calculation is correct. Wait, but let's confirm the z-table. For z = 0.4, the cumulative probability is 0.6554 (since z=0.40 gives 0.6554). So $P(z > 0.4) = 1 - 0.6554 = 0.3446$. Wait, but the options have 0.3446 as one of them. Wait, but let's check again. Wait, maybe I made a mistake in the z-score. Wait, $\sigma_{\bar{x}} = 12 / \sqrt{36} = 2$, correct. $\bar{x} - \mu = 0.8$, so z = 0.8 / 2 = 0.4, correct. Then $P(\bar{x} > 50.8) = P(z > 0.4) = 1 - 0.6554 = 0.3446$. So the answer is 0.3446.

Wait, but let me check again. Wait, maybe the question is about the sampling distribution of the sample mean, which is normal (by CLT, since n=36 is large). So the calculation is correct. So the probability is 0.3446.

Step1: Recall the formula for the confidence interval

For a 95% confidence interval when the population standard deviation ($\sigma$) is known, the formula is:
$\bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}$

Step2: Identify the values

Given:

• Sample mean ($\bar{x}$) = 1883 KWH
• Population standard deviation ($\sigma$) = 480 KWH
• Sample size ($n$) = 100
• For 95% confidence, $z_{\alpha/2} = 1.96$ (since $\alpha = 0.05$, so $\alpha/2 = 0.025$, and $z_{0.025} = 1.96$)

Step3: Calculate the margin of error (E)

$E = z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}$
$E = 1.96 \cdot \frac{480}{\sqrt{100}}$
$\sqrt{100} = 10$, so:
$E = 1.96 \cdot \frac{480}{10} = 1.96 \cdot 48 = 94.08$

Step4: Calculate the confidence interval

Lower limit: $\bar{x} - E = 1883 - 94.08 = 1788.92$
Upper limit: $\bar{x} + E = 1883 + 94.08 = 1977.08$

So the 95% confidence interval is (1788.92, 1977.08).

Answer:

0.3446 (corresponding to the option with 0.3446)

Question 10