We are dealing with a normally distributed set of test scores with the following parameters:

• Mean ($\mu$) = 22 points
• Standard deviation ($\sigma$) = 2 points

(a) Proportion of students who scored at least 23 points:

To find this proportion, we first standardize the score of 23 points using the Z-score formula:

$Z = \frac{X - \mu}{\sigma}$

Where:

• $X = 23$
• $\mu = 22$
• $\sigma = 2$

$Z = \frac{23 - 22}{2} = \frac{1}{2} = 0.5$

Next, we find the cumulative probability for a Z-score of 0.5 using the standard normal distribution table or a calculator. The cumulative probability for $Z = 0.5$ is approximately 0.69146.

This value represents the proportion of students who scored less than 23 points. To find the proportion of students who scored at least 23 points, we subtract this value from 1:

$P(X \geq 23) = 1 - P(Z \leq 0.5) = 1 - 0.69146 = 0.30854$

Thus, the proportion of students who scored at least 23 points is 0.30854 (rounded to five decimal places).

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(b) The 20th percentile of the distribution of test scores:

The 20th percentile corresponds to the score below which 20% of the students fall. First, we find the Z-score that corresponds to the 20th percentile ($P = 0.20$) using a standard normal distribution table or a calculator.

The Z-score for the 20th percentile is approximately -0.8416.

Now, we convert this Z-score back to the original score using the Z-score formula:

$X = \mu + Z\sigma$

Substitute the known values:

$X = 22 + (-0.8416)(2) = 22 - 1.6832 = 20.3168$

Thus, the 20th percentile of the test scores is approximately 20.317 (rounded to three decimal places).

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Summary:

• (a) The proportion of students who scored at least 23 points is 0.30854.
• (b) The 20th percentile of the distribution is 20.317.

Would you like further clarification or more details on any part of this explanation?

Here are some related questions:

1. How would the result change if the mean were different?
2. What proportion of students scored less than 21 points?
3. What is the 90th percentile of this test distribution?
4. If the standard deviation was larger, how would the percentile calculations change?
5. What would the Z-score be for a student who scored 24 points?

Tip: Understanding the Z-score helps you compare values from different normal distributions efficiently!