The heights of a group of 400 third graders are normally distributed with a mean of 52 inches and a standard deviation of 1.6 inches. What height would a third grader need to be in order to be in the top 2.5% tallest third graders in this group? Give your answer to 1 decimal place. Do not include the word inches in your answer
The Correct Answer and Explanation is:
To determine the height required to be in the top 2.5% of third graders, we use the properties of the normal distribution and find the corresponding z-score.
Step 1: Identify given values Mean (μ) = 52 Standard deviation (σ) = 1.6 Top 2.5% corresponds to the 97.5th percentile in a normal distribution.
Step 2: Find the z-score From statistical tables, the z-score corresponding to the 97.5th percentile is approximately 1.96.
Step 3: Use the z-score formula
X=μ+zσX = \mu + z\sigma
Substituting values,
X=52+(1.96×1.6)X = 52 + (1.96 \times 1.6)
X=52+3.136X = 52 + 3.136
X=55.1X = 55.1
Thus, a third grader must be at least 55.1 to be in the top 2.5% of the group.
Explanation
The normal distribution is a symmetric probability function that describes how values cluster around the mean. In this case, student heights follow a bell-shaped curve centered at 52. The standard deviation of 1.6 indicates how spread out the heights are. The larger the standard deviation, the wider the spread.
To find the cutoff height for the tallest 2.5%, we use the z-score, which measures how many standard deviations a value is from the mean. Using statistical tables, we determined that a z-score of 1.96 corresponds to the 97.5th percentile. Plugging into the formula, we found that students taller than 55.1 are among the tallest 2.5%. This statistical method ensures an accurate cutoff point based on probability theory.
