Questiol C7 A distribution of values is normal with a mean of 8.9 and a standard deviation of 66.1. Find the probability that a randomly selected value is greater than -196. P(X > -196) = Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or scores rounded to 3 decimal places are accepted. Submit Question

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**Statistics and Probability: Working with Normal Distributions**

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**Question 25**
   
A distribution of values is normal with a mean of 8.9 and a standard deviation of 66.1.

Find the probability that a randomly selected value is greater than -196.

\[ P(X > -196) = \boxed{\phantom{number}} \]

Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

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**Instructions for Answering:**

1. Calculate the z-score for the value of -196 using the formula:

\[ z = \frac{X - \mu}{\sigma} \]

   where:
   
   - \( X \) is the value (-196 in this case)
   - \( \mu \) is the mean (8.9 in this case)
   - \( \sigma \) is the standard deviation (66.1 in this case)

2. Use the z-score to find the probability from the standard normal distribution table or using a calculator with statistical functions.

3. Enter your answer as a decimal rounded to 4 decimal places.

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**Submit Your Answer:**

Click the "Submit Question" button to record your answer once you have calculated the probability.

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**Example Text Explanation:**

1. Calculate the z-score:

\[ z = \frac{-196 - 8.9}{66.1} \approx -3.1097 \]

2. Find the probability corresponding to the z-score \(-3.1097\):
   
   From the standard normal distribution table, the probability (P) corresponding to \( z \approx -3.11 \) is approximately \( 0.0009 \).

3. Since we want the probability \( P(X > -196) \):

\[ P(X > -196) = 1 - P(X \leq -196) = 1 - 0.0009 = 0.9991 \]

4. Finally, enter \( 0.9991 \) as your answer in the provided field.

---

**Reminder:**

Ensure your answers are precise and rounded as required. This exercise helps in understanding the application of the normal distribution and z-scores in determining probabilities.
Transcribed Image Text:**Statistics and Probability: Working with Normal Distributions** --- **Question 25** A distribution of values is normal with a mean of 8.9 and a standard deviation of 66.1. Find the probability that a randomly selected value is greater than -196. \[ P(X > -196) = \boxed{\phantom{number}} \] Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted. --- **Instructions for Answering:** 1. Calculate the z-score for the value of -196 using the formula: \[ z = \frac{X - \mu}{\sigma} \] where: - \( X \) is the value (-196 in this case) - \( \mu \) is the mean (8.9 in this case) - \( \sigma \) is the standard deviation (66.1 in this case) 2. Use the z-score to find the probability from the standard normal distribution table or using a calculator with statistical functions. 3. Enter your answer as a decimal rounded to 4 decimal places. --- **Submit Your Answer:** Click the "Submit Question" button to record your answer once you have calculated the probability. --- **Example Text Explanation:** 1. Calculate the z-score: \[ z = \frac{-196 - 8.9}{66.1} \approx -3.1097 \] 2. Find the probability corresponding to the z-score \(-3.1097\): From the standard normal distribution table, the probability (P) corresponding to \( z \approx -3.11 \) is approximately \( 0.0009 \). 3. Since we want the probability \( P(X > -196) \): \[ P(X > -196) = 1 - P(X \leq -196) = 1 - 0.0009 = 0.9991 \] 4. Finally, enter \( 0.9991 \) as your answer in the provided field. --- **Reminder:** Ensure your answers are precise and rounded as required. This exercise helps in understanding the application of the normal distribution and z-scores in determining probabilities.
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