
MATLAB: An Introduction with Applications
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ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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![### Probability of Mean Test Scores Less Than 79 in Seaturtle Research Studies
A final exam in Seaturtle Research Studies has a mean of 78 and a standard deviation of 3. If 36 students are randomly selected, we aim to find the probability that the mean of their test scores is less than 79.
#### Question
Given:
- Mean (\(\mu\)) = 78
- Standard deviation (\(\sigma\)) = 3
- Number of students (n) = 36
Find the probability \(P(x < 79)\):
\[ P(x < 79) = P\left(z < \frac{79 - \mu}{\frac{\sigma}{\sqrt{n}}} \right) = \]
\[ P\left(z < \frac{79 - 78}{\frac{3}{\sqrt{36}}}\right) = P\left(z < \frac{1}{\frac{3}{6}}\right) = P(z < 2) \]
Illustrate and calculate this probability using the standard normal distribution:
#### Standard Normal Distribution Explanation
Below is a standard normal distribution graph used to visually illustrate the probability:

- **Shaded Area**: The region to the left of \(z = 2\) is shaded in blue.
- **Meaning**: The shaded area represents \(P(z < 2)\).
- The arrow indicates the z-score of -1.5 corresponding to the 79 marks based on the standard normal distribution.
#### Instructions
Enter your answer as a percentage accurate to 1 decimal place without the "%" sign.
\[ P(z < 2) = \]
Let's calculate the exact percentage and fill in the box:
\[ P(z < 2) \approx 97.7\% \]
Thus:
\[ P(x < 79) = 97.7 \]
Enter the value:
\[ \boxed{97.7} \]
Remember to use this calculator:
Shade: Left of a value: \(87.7\)
Click and drag the arrows to adjust the values to calculate the precise z-score related probabilities.](https://content.bartleby.com/qna-images/question/f21c709c-710e-425a-a9e1-f3b584be2fe0/c5da6f19-7e3a-4ebf-a1d0-ce22606c288b/ivvgtke_thumbnail.png)
Transcribed Image Text:### Probability of Mean Test Scores Less Than 79 in Seaturtle Research Studies
A final exam in Seaturtle Research Studies has a mean of 78 and a standard deviation of 3. If 36 students are randomly selected, we aim to find the probability that the mean of their test scores is less than 79.
#### Question
Given:
- Mean (\(\mu\)) = 78
- Standard deviation (\(\sigma\)) = 3
- Number of students (n) = 36
Find the probability \(P(x < 79)\):
\[ P(x < 79) = P\left(z < \frac{79 - \mu}{\frac{\sigma}{\sqrt{n}}} \right) = \]
\[ P\left(z < \frac{79 - 78}{\frac{3}{\sqrt{36}}}\right) = P\left(z < \frac{1}{\frac{3}{6}}\right) = P(z < 2) \]
Illustrate and calculate this probability using the standard normal distribution:
#### Standard Normal Distribution Explanation
Below is a standard normal distribution graph used to visually illustrate the probability:

- **Shaded Area**: The region to the left of \(z = 2\) is shaded in blue.
- **Meaning**: The shaded area represents \(P(z < 2)\).
- The arrow indicates the z-score of -1.5 corresponding to the 79 marks based on the standard normal distribution.
#### Instructions
Enter your answer as a percentage accurate to 1 decimal place without the "%" sign.
\[ P(z < 2) = \]
Let's calculate the exact percentage and fill in the box:
\[ P(z < 2) \approx 97.7\% \]
Thus:
\[ P(x < 79) = 97.7 \]
Enter the value:
\[ \boxed{97.7} \]
Remember to use this calculator:
Shade: Left of a value: \(87.7\)
Click and drag the arrows to adjust the values to calculate the precise z-score related probabilities.
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