eat the same cookie snack every night. The amount of aturated fat in the serving is normally distributed with a hean of 26 grams and a standard deviation of 3 grams. ind the probability that my saturated fat intake is below 7 grams? Would this event be considered unusual? xplain your reasoning.
eat the same cookie snack every night. The amount of aturated fat in the serving is normally distributed with a hean of 26 grams and a standard deviation of 3 grams. ind the probability that my saturated fat intake is below 7 grams? Would this event be considered unusual? xplain your reasoning.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Question
![**Probability and Statistics Example: Calculating Probabilities and Understanding Unusual Events**
**Problem Description:**
I eat the same cookie snack every night. The amount of saturated fat in the serving is normally distributed with a mean of 26 grams and a standard deviation of 3 grams. Find the probability that my saturated fat intake is below 27 grams? Would this event be considered unusual? Explain your reasoning.
**Solution Explanation:**
To solve this problem, we need to use the properties of the normal distribution.
**Step-by-Step Solution:**
1. **Identify the Given Parameters:**
- Mean (μ) = 26 grams
- Standard Deviation (σ) = 3 grams
- We need to find the probability that the saturated fat intake (X) is below 27 grams, i.e., P(X < 27).
2. **Standardize the Normal Variable:**
To find this probability, we need to convert the value of 27 grams to its corresponding z-score in the standard normal distribution.
The formula for the z-score is:
\[
z = \frac{X - \mu}{\sigma}
\]
Plugging in the values:
\[
z = \frac{27 - 26}{3} = \frac{1}{3} \approx 0.33
\]
3. **Find the Probability Using the Z-Table:**
Using the z-table (standard normal distribution table), we look up the z-score of 0.33. The z-table gives the probability to the left of z.
- The z-table value for 0.33 is approximately 0.6293.
- This means P(X < 27) ≈ 0.6293.
4. **Determine if the Event is Unusual:**
- An event is generally considered unusual if its probability is less than 0.05 (or 5%).
- In this case, the probability we found is approximately 0.6293, which is much greater than 0.05.
- Therefore, having a saturated fat intake below 27 grams is not considered unusual.
**Conclusion:**
The probability that the saturated fat intake is below 27 grams is approximately 0.6293, or 62.93%. Since this probability is much higher than 5%, this event is not considered unusual.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb31d7e90-5e73-4234-85af-d540cd23e9cc%2Fe45d95d2-90c9-4cf4-b969-5a3145774462%2F4broy7d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Probability and Statistics Example: Calculating Probabilities and Understanding Unusual Events**
**Problem Description:**
I eat the same cookie snack every night. The amount of saturated fat in the serving is normally distributed with a mean of 26 grams and a standard deviation of 3 grams. Find the probability that my saturated fat intake is below 27 grams? Would this event be considered unusual? Explain your reasoning.
**Solution Explanation:**
To solve this problem, we need to use the properties of the normal distribution.
**Step-by-Step Solution:**
1. **Identify the Given Parameters:**
- Mean (μ) = 26 grams
- Standard Deviation (σ) = 3 grams
- We need to find the probability that the saturated fat intake (X) is below 27 grams, i.e., P(X < 27).
2. **Standardize the Normal Variable:**
To find this probability, we need to convert the value of 27 grams to its corresponding z-score in the standard normal distribution.
The formula for the z-score is:
\[
z = \frac{X - \mu}{\sigma}
\]
Plugging in the values:
\[
z = \frac{27 - 26}{3} = \frac{1}{3} \approx 0.33
\]
3. **Find the Probability Using the Z-Table:**
Using the z-table (standard normal distribution table), we look up the z-score of 0.33. The z-table gives the probability to the left of z.
- The z-table value for 0.33 is approximately 0.6293.
- This means P(X < 27) ≈ 0.6293.
4. **Determine if the Event is Unusual:**
- An event is generally considered unusual if its probability is less than 0.05 (or 5%).
- In this case, the probability we found is approximately 0.6293, which is much greater than 0.05.
- Therefore, having a saturated fat intake below 27 grams is not considered unusual.
**Conclusion:**
The probability that the saturated fat intake is below 27 grams is approximately 0.6293, or 62.93%. Since this probability is much higher than 5%, this event is not considered unusual.
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