Question
The scores of a high :chool entrance exam are approximately normally distributed with a given mean \( \mu=82.4 \) and standard deviation \( \sigma=3.3 \). What percentage of the scores are between 75.8 and 89 ? \( 68 \% \) \( 95 \% \) \( 99.7 \% \) \( 100 \% \)
Ask by Hill Pollard. in the United States
Feb 28,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
About 95% of the scores are between 75.8 and 89.
Solution
To solve this problem, we first standardize the interval. The given normal distribution has a mean μ = 82.4 and standard deviation σ = 3.3. We need to find the percentage of scores between 75.8 and 89.
Step 1: Convert the raw scores to z-scores.
For the lower bound (75.8):
z = (75.8 – 82.4) / 3.3 = –6.6 / 3.3 = –2
For the upper bound (89):
z = (89 – 82.4) / 3.3 = 6.6 / 3.3 = 2
Step 2: Identify the percentage of observations between z = –2 and z = 2.
According to the empirical rule (68-95-99.7 rule) for a normal distribution, approximately 95% of the data fall within 2 standard deviations of the mean.
Thus, about 95% of the scores are between 75.8 and 89.
The correct answer is 95%.
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Simplify this solution Beyond the Answer
To solve this problem, we can use the properties of the normal distribution. First, convert the raw scores to z-scores. For the low score (75.8): \[ z = \frac{75.8 - 82.4}{3.3} \approx -2 \] For the high score (89): \[ z = \frac{89 - 82.4}{3.3} \approx 2 \] Now, using the empirical rule (or the 68-95-99.7 rule), we can deduce that approximately 95% of the data falls within \(-2\) and \(+2\) standard deviations from the mean. Hence, roughly 95% of the scores are between 75.8 and 89. The answer is \( 95 \% \).
