Heights (X)  of college students  are known to follow an approximately  bell-shaped symmetric distribution with mean
= 171.3 cm and standard deviation
= 4.1 cm.  Suppose your height is the simplified value h = (171.3 + 4.1) cm.  Approximately, what percentage of students are expected to be taller than you?   Enter the answer without the % sign (e.g. 40% should be entered as 40, 40.6% should be entered as 40.6). Report answer up to two decimal places without rounding up or down.

Given the information:
- Mean (μ) = 171.3 cm
- Standard deviation (σ) = 4.1 cm
- Your height (h) = 171.3 + 4.1 = 175.4 cm

To find the percentage of students taller than you, we need to calculate the z-score and then use the standard normal distribution to find the corresponding percentile.

The z-score is calculated as:
\[
z = \frac{h - \mu}{\sigma}
\]

Substituting the values:
\[
z = \frac{175.4 - 171.3}{4.1} = \frac{4.1}{4.1} = 1
\]

A z-score of 1 corresponds to the 84.13th percentile of the standard normal distribution. This means that 84.13% of students are shorter than 175.4 cm.

To find the percentage of students taller than you:
\[
100\% - 84.13\% = 15.87\%
\]

Thus, approximately 15.87% of students are expected to be taller than you. 

The answer is:

\[ 15.87 \]

Would you like more details or have any questions? Here are some related questions you might find useful:

1. What is the z-score and how is it used in statistics?
2. How do you interpret the results of a z-score?
3. What does the standard normal distribution represent?
4. Can you explain the concept of percentiles in a normal distribution?
5. How is the standard deviation used in calculating z-scores?
6. What is the empirical rule (68-95-99.7 rule) in a normal distribution?
7. How can you use a z-table to find probabilities and percentiles?
8. What are some applications of z-scores in real life?

**Tip:** When working with the normal distribution, it's helpful to remember key z-score values and their associated percentiles for quick reference. For example, z = 1.0 is approximately the 84th percentile, z = 2.0 is approximately the 97.5th percentile, and z = -1.0 is approximately the 16th percentile.

Learn how to calculate the percentage of students expected to be taller than a given height in a normal distribution. This problem involves using the mean, standard deviation, and z-score formula to find the percentile in a bell-shaped symmetric distribution.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation