Assume a member is selected at random from the population represented by the graph. Find the probability that the member selected at random is from the shaded region of the graph. Assume the variable x is normally distributed.
Standardized Test Composite Scores
26
6
32
Score
mu equals 19.9sigma equals 5.426 less than x less than 32
x



A graph titled "Standardized Test Composite Scores" has a horizontal x-axis labeled "Score" from about 1 to 39 with tick marks at 6, 26, and 32. A normal curve labeled μ = 19.9 σ = 5.4 is centered above the x-axis at 19.9. Two vertical line segments extend from the curve to the x-axis at 26 and 32. The area below the curve and between the two vertical line segments is shaded and labeled 26 < x < 32.
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Part 1
The probability that the member selected at random is from the shaded area of the graph is 
  
enter your response here. ​(Round to four decimal places as​ needed.)

To find the probability that a member selected at random has a score between 26 and 32 from a normal distribution, we need to calculate the probability that \( 26 < x < 32 \) for the given normal distribution.

### Given information:
- Mean (\( \mu \)) = 19.9
- Standard deviation (\( \sigma \)) = 5.4
- We want to find \( P(26 < x < 32) \).

To do this, we can follow these steps:

### Step 1: Convert the raw scores to z-scores
The z-score is calculated using the formula:

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

1. **For \( x = 26 \):**
   \[
   z_1 = \frac{26 - 19.9}{5.4} = \frac{6.1}{5.4} \approx 1.13
   \]

2. **For \( x = 32 \):**
   \[
   z_2 = \frac{32 - 19.9}{5.4} = \frac{12.1}{5.4} \approx 2.24
   \]

### Step 2: Find the cumulative probabilities
We use standard normal distribution tables or a calculator to find the cumulative probabilities for these z-scores.

- \( P(z_1 = 1.13) \approx 0.8708 \)
- \( P(z_2 = 2.24) \approx 0.9875 \)

### Step 3: Calculate the probability between the two z-scores
The probability that \( 26 < x < 32 \) is the difference between the cumulative probabilities for \( z_2 \) and \( z_1 \):

\[
P(26 < x < 32) = P(z_2) - P(z_1) = 0.9875 - 0.8708 = 0.1167
\]

### Final Answer:
The probability that a randomly selected member has a score between 26 and 32 is approximately **0.1167**.

Would you like more details or have any questions on this? 

Here are 5 related questions to deepen understanding:
1. How is the z-score used in normal distributions?
2. How do you interpret the value of the cumulative probability in a normal distribution?
3. How would the probability change if the standard deviation were smaller?
4. What is the significance of finding probabilities between two points on a normal distribution?
5. How does increasing the mean affect the shaded region in a normal distribution?

**Tip:** When using z-scores, always double-check that the data follows a normal distribution, as this method relies on that assumption.

Assume a member is selected at random from the population represented by the graph. Find the probability that the member selected at random is from the shaded region of the graph. Assume the variable x is normally distributed. Standardized Test Composite Scores: 26, 6, 32. Score μ = 19.9, σ = 5.4. 26 < x < 32.

Learn how to find the probability that a standardized test score falls between 26 and 32, assuming a normal distribution with mean μ = 19.9 and standard deviation σ = 5.4. This problem demonstrates how to use z-scores and cumulative probabilities in normal distributions. Ideal for Grades 10-12.

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