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Let's break down and solve each part of the problem from the image:

### Problem Information:
The aptitude test scores of applicants are normally distributed with:
- **Mean (μ)** = 400
- **Standard Deviation (σ)** = 100

### Part a:
An applicant needs a test score higher than 525 to be admitted to the graduate program. What proportion of candidates qualifies?

To solve this, we will use the **Z-score formula**:
\[
Z = \frac{X - \mu}{\sigma}
\]
Where:
- \( X \) = 525 (the score)
- \( \mu \) = 400 (the mean)
- \( \sigma \) = 100 (the standard deviation)

\[
Z = \frac{525 - 400}{100} = \frac{125}{100} = 1.25
\]

Now, we need to find the proportion of the normal distribution that corresponds to a Z-score of 1.25. Using a Z-table or a standard normal distribution calculator:
- The area to the left of Z = 1.25 is approximately 0.8944 (or 89.44%).

Since the problem asks for the proportion of candidates who score higher than 525, we need to find the area to the right of Z = 1.25:
\[
P(X > 525) = 1 - 0.8944 = 0.1056
\]

Thus, about **10.56% of candidates qualify**.

---

### Part b:
The university wishes to set the cut-off score for graduate admission so that only the top 10 percent of applicants qualify. What is the required cut-off score?

For this, we need to find the Z-score that corresponds to the top 10% of the distribution. From the Z-table, the Z-score for the 90th percentile (since 90% are below the score) is approximately **Z = 1.28**.

Now, use the Z-score formula to find the cut-off score \(X\):
\[
Z = \frac{X - \mu}{\sigma}
\]
Substitute the known values:
\[
1.28 = \frac{X - 400}{100}
\]
Solve for \(X\):
\[
X = 1.28 \times 100 + 400 = 128 + 400 = 528
\]

The required cut-off score is **528**.

---

### Part c:
What percentage of applicants has test scores within two standard deviations of the mean?

Two standard deviations from the mean corresponds to:
\[
\mu - 2\sigma \leq X \leq \mu + 2\sigma
\]
\[
400 - 2(100) \leq X \leq 400 + 2(100)
\]
\[
200 \leq X \leq 600
\]

We need to find the proportion of the distribution between these two scores. The Z-scores corresponding to 200 and 600 are:
- For \(X = 200\): \(Z = \frac{200 - 400}{100} = -2\)
- For \(X = 600\): \(Z = \frac{600 - 400}{100} = 2\)

From the Z-table:
- The area to the left of Z = -2 is approximately 0.0228 (2.28%).
- The area to the left of Z = 2 is approximately 0.9772 (97.72%).

The percentage of applicants within two standard deviations is:
\[
P(200 \leq X \leq 600) = 0.9772 - 0.0228 = 0.9544
\]

Thus, about **95.44%** of applicants have test scores within two standard deviations of the mean.

---

Do you need further details on any part of this problem?

### Follow-up Questions:
1. How would the results change if the standard deviation were larger or smaller?
2. How can we interpret the top 5% cut-off score instead of the top 10%?
3. What would happen to the proportion of candidates qualifying if the required score was raised to 550?
4. How do the areas under the normal curve help in setting admission criteria?
5. Can we apply the same method to non-normal distributions? Why or why not?

### Tip:
When using Z-tables or normal distribution tables, always keep track of whether you are looking for areas to the left or right of the Z-score!

The aptitude test scores of applicants to a university graduate program are normally distributed with mean = 400 and standard deviation = 100. The problem asks: a) What proportion of candidates qualifies with a score higher than 525? b) What cut-off score would place applicants in the top 10%? c) What percentage of applicants has test scores within two standard deviations of the mean?

Solve a problem involving normally distributed test scores with a mean of 400 and standard deviation of 100. Learn how to find the proportion of applicants qualifying with scores above 525, the cut-off score for the top 10%, and the percentage of applicants within two standard deviations of the mean.

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