Question

In an all boys school, the heights of the student body are normally distributed with a mean of 71 inches and a standard deviation of 3 inches. What percentage of the students are between 72 and 79 inches tall, to the nearest tenth?

Answer 1

To find the percentage of students whose heights are between 72 and 79 inches in a normally distributed population, we will use the properties of the normal distribution.

### Step 1: Standardize the Heights
We will convert the heights of 72 and 79 inches into z-scores using the formula:

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

where:
- \( \mu = 71 \) (mean)
- \( \sigma = 3 \) (standard deviation)

#### For 72 inches:
\[
z_{72} = \frac{72 - 71}{3} = \frac{1}{3} \approx 0.333
\]

#### For 79 inches:
\[
z_{79} = \frac{79 - 71}{3} = \frac{8}{3} \approx 2.667
\]

### Step 2: Use Z-Table (or Calculator) to Find Percentages
Next, we will find the area under the standard normal distribution curve corresponding to these z-scores.

Using a standard normal distribution table or calculator:

1. **For \( z_{72} = 0.333 \)**:
- The cumulative probability (area to the left) is approximately \( 0.6293 \).

2. **For \( z_{79} = 2.667 \)**:
- The cumulative probability (area to the left) is approximately \( 0.9963 \).

### Step 3: Calculate the Area Between the Two Z-Scores
To find the proportion of students between 72 inches and 79 inches, we subtract the cumulative probabilities:

\[
P(72 < X < 79) = P(X < 79) - P(X < 72)
\]
\[
P(72 < X < 79) = 0.9963 - 0.6293 = 0.3670
\]

### Step 4: Convert to Percentage
To express this as a percentage, we multiply by 100:

\[
\text{Percentage} = 0.3670 \times 100 \approx 36.7\%
\]

### Conclusion
The percentage of students who are between 72 and 79 inches tall is approximately **36.7%**.