The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 73 inches, and a standard deviation of 10 inches. What is the probability that the mean annual snowfall during 25 randomly picked years will exceed 75.8 inches? Round your answer to four decimal places. OA. 0.0026 OB. 0.0808 OC. 0.5808 OD. 0.4192
The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 73 inches, and a standard deviation of 10 inches. What is the probability that the mean annual snowfall during 25 randomly picked years will exceed 75.8 inches? Round your answer to four decimal places. OA. 0.0026 OB. 0.0808 OC. 0.5808 OD. 0.4192
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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![### Understanding Probability in Normally Distributed Snowfall Data
#### Question:
The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 73 inches, and a standard deviation of 10 inches. What is the probability that the mean annual snowfall during 25 randomly picked years will exceed 75.8 inches? Round your answer to four decimal places.
#### Answer Choices:
- A. 0.0026
- B. 0.0808
- C. 0.5808
- D. 0.4192
#### Explanation:
To solve this problem, we need to find the probability that the sample mean of annual snowfall for 25 years exceeds 75.8 inches, given that the snowfall follows a normal distribution with a mean (\(\mu\)) of 73 inches and a standard deviation (\(\sigma\)) of 10 inches.
The formula for the standard error of the mean (SEM) when dealing with a sample from a normal distribution is:
\[ \text{SEM} = \frac{\sigma}{\sqrt{n}} \]
Where \(\sigma\) is the standard deviation and \(n\) is the sample size.
Given:
- \(\mu = 73 \, \text{inches}\)
- \(\sigma = 10 \, \text{inches}\)
- \(n = 25\)
Calculate the standard error of the mean (SEM):
\[ \text{SEM} = \frac{10}{\sqrt{25}} = \frac{10}{5} = 2 \]
Next, convert the sample mean of 75.8 inches to a z-score using the formula:
\[ z = \frac{\bar{x} - \mu}{\text{SEM}} \]
Where \(\bar{x}\) is the sample mean.
\[ z = \frac{75.8 - 73}{2} = \frac{2.8}{2} = 1.4 \]
Using the z-score table (Standard Normal Distribution table), we can find the probability corresponding to a z-score of 1.4. This probability represents the area under the curve to the left of the z-score.
- The area to the left of \(z = 1.4\) is approximately 0.9192.
The probability of exceeding 75.8 inches of snowfall (\(P\)) is:
\[ P(\bar{x} >](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4b33feb0-e62b-4e25-aad1-2d23e491b0f1%2Fe2821c4c-8528-49b7-93f3-91429b96e717%2Ftkux0p5_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Probability in Normally Distributed Snowfall Data
#### Question:
The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 73 inches, and a standard deviation of 10 inches. What is the probability that the mean annual snowfall during 25 randomly picked years will exceed 75.8 inches? Round your answer to four decimal places.
#### Answer Choices:
- A. 0.0026
- B. 0.0808
- C. 0.5808
- D. 0.4192
#### Explanation:
To solve this problem, we need to find the probability that the sample mean of annual snowfall for 25 years exceeds 75.8 inches, given that the snowfall follows a normal distribution with a mean (\(\mu\)) of 73 inches and a standard deviation (\(\sigma\)) of 10 inches.
The formula for the standard error of the mean (SEM) when dealing with a sample from a normal distribution is:
\[ \text{SEM} = \frac{\sigma}{\sqrt{n}} \]
Where \(\sigma\) is the standard deviation and \(n\) is the sample size.
Given:
- \(\mu = 73 \, \text{inches}\)
- \(\sigma = 10 \, \text{inches}\)
- \(n = 25\)
Calculate the standard error of the mean (SEM):
\[ \text{SEM} = \frac{10}{\sqrt{25}} = \frac{10}{5} = 2 \]
Next, convert the sample mean of 75.8 inches to a z-score using the formula:
\[ z = \frac{\bar{x} - \mu}{\text{SEM}} \]
Where \(\bar{x}\) is the sample mean.
\[ z = \frac{75.8 - 73}{2} = \frac{2.8}{2} = 1.4 \]
Using the z-score table (Standard Normal Distribution table), we can find the probability corresponding to a z-score of 1.4. This probability represents the area under the curve to the left of the z-score.
- The area to the left of \(z = 1.4\) is approximately 0.9192.
The probability of exceeding 75.8 inches of snowfall (\(P\)) is:
\[ P(\bar{x} >
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