A set of elementary school student heights are normally distributed with a mean of 
\[105\] centimeters and a standard deviation of 
\[7\] centimeters.
What proportion of student heights are between 
\[94.5\] centimeters and 
\[115.5\] centimeters?
You may round your answer to four decimal places.

To find the proportion of student heights between \( 94.5 \) centimeters and \( 115.5 \) centimeters, we'll use the properties of the normal distribution.

### Step 1: Standardize the Values
We will convert the heights \( 94.5 \) cm and \( 115.5 \) cm to their corresponding z-scores using the formula:

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

Where:
- \( x \) is the value we want to standardize,
- \( \mu = 105 \) cm is the mean,
- \( \sigma = 7 \) cm is the standard deviation.

#### For \( x = 94.5 \) cm:
\[
z_1 = \frac{94.5 - 105}{7} = \frac{-10.5}{7} \approx -1.5
\]

#### For \( x = 115.5 \) cm:
\[
z_2 = \frac{115.5 - 105}{7} = \frac{10.5}{7} \approx 1.5
\]

### Step 2: Find the Corresponding Probabilities
Next, we find the probabilities corresponding to these z-scores using the standard normal distribution table or a calculator.

#### Probability for \( z_1 = -1.5 \):
\[
P(Z \leq -1.5) \approx 0.0668
\]

#### Probability for \( z_2 = 1.5 \):
\[
P(Z \leq 1.5) \approx 0.9332
\]

### Step 3: Calculate the Proportion
The proportion of student heights between \( 94.5 \) cm and \( 115.5 \) cm is the difference between these two probabilities:

\[
P(94.5 \leq X \leq 115.5) = P(Z \leq 1.5) - P(Z \leq -1.5) = 0.9332 - 0.0668 = 0.8664
\]

### Final Answer:
The proportion of student heights between \( 94.5 \) cm and \( 115.5 \) cm is approximately \( \boxed{0.8664} \).

Would you like more details, or do you have any questions?

Here are 5 related questions:
1. What proportion of student heights are less than 94.5 cm?
2. What proportion of student heights are greater than 115.5 cm?
3. What proportion of student heights are between 98 cm and 112 cm?
4. How would the proportion change if the standard deviation were 10 cm instead of 7 cm?
5. How can we use the z-score to find the height corresponding to the 90th percentile?

**Tip:** Remember that z-scores can help compare data points from different normal distributions by standardizing them.

Explore how to calculate the proportion of student heights between 94.5 cm and 115.5 cm in a normal distribution with mean 105 cm and standard deviation 7 cm. Learn about Z-scores, probability calculations, and the properties of normal distribution. Suitable for Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation