Question

a meteorological organization is comparing annual snowfall patterns in two cities, brightown and snowyvalle. the cities have the following characteristics: table with city (brightown, snowyvalle), mean annual snowfall (inches) (23, 41), standard deviation (inches) (3, 3). a sample of the annual snowfall for each city is collected over 34 years in brightown and over 47 years in snowyvalle. a. describe the probability distribution of the sample mean annual snowfall for brightown. (round numerical answers to four decimal places.) the sample mean annual snowfall for brightown follows a inches and $sigma_{\bar{x}}=$ inches. distribution with $e(\bar{x})=$ b. find the probability that brightowns sample mean is within 0.9 inches of its population mean. (round your answer to four decimal places.) c. find the probability that snowyvalles sample mean is within 0.9 inches of its population mean. (round your answer to four decimal places.)

Explanation:

Step1: Recall the formula for the mean of a sample

The sample - mean $\bar{x}$ is an unbiased estimator of the population mean $\mu$. For a sample of size $n$, the mean of the sampling distribution of the sample mean $\mu_{\bar{x}}=\mu$ and the standard deviation of the sampling distribution of the sample mean (also known as the standard error) is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$.

Step2: Calculate the standard error for Brightown

For Brightown, $n = 34$, $\sigma=3$. The standard error $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{34}}\approx\frac{3}{5.831}\approx0.515$.

Step3: Calculate the z - score for the margin of error

We want to find the probability that the sample mean is within $0.9$ inches of the population mean. The z - score is calculated as $z=\frac{\text{Margin of Error}}{\sigma_{\bar{x}}}$. Here, the margin of error $E = 0.9$. So, $z=\frac{0.9}{0.515}\approx1.75$.

Step4: Use the standard normal distribution table

The probability that $- 1.75

Step5: Repeat the process for Snowyvalle

For Snowyvalle, $n = 47$, $\sigma = 3$. The standard error $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{47}}\approx\frac{3}{6.856}\approx0.438$.
The z - score for a margin of error of $E = 0.9$ is $z=\frac{0.9}{0.438}\approx2.05$.
The probability that $-2.05

a. For Brightown:
The sampling distribution of the sample mean annual snowfall is approximately normal with mean $\mu_{\bar{x}}=\mu$ (the population mean) and standard deviation (standard error) $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{34}}\approx0.515$ inches.
For Snowyvalle:
The sampling distribution of the sample mean annual snowfall is approximately normal with mean $\mu_{\bar{x}}=\mu$ (the population mean) and standard deviation (standard error) $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{47}}\approx0.438$ inches.
b. For Brightown, the probability that the sample mean is within $0.9$ inches of its population mean is approximately $0.9198$.
c. For Snowyvalle, the probability that the sample mean is within $0.9$ inches of its population mean is approximately $0.9600$.

Answer:

a. Brightown: Sampling distribution of $\bar{x}$ is normal with $\mu_{\bar{x}}=\mu$ (population mean) and $\sigma_{\bar{x}}\approx0.515$ inches. Snowyvalle: Sampling distribution of $\bar{x}$ is normal with $\mu_{\bar{x}}=\mu$ and $\sigma_{\bar{x}}\approx0.438$ inches.
b. $0.9198$
c. $0.9600$