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.