Question

on a nationwide test taken by high school students, the mean score was 47 and the standard deviation was 13. the scores were normally distributed. complete the following statements.
(a) approximately 7 of the students scored between 34 and 60.
(b) approximately 99.7% of the students scored between and .

Explanation:

Step1: Identify z-scores for part (a)

First, calculate z-scores for 34 and 60 using $z=\frac{x-\mu}{\sigma}$, where $\mu=47$, $\sigma=13$.
For $x=34$: $z=\frac{34-47}{13}=\frac{-13}{13}=-1$
For $x=60$: $z=\frac{60-47}{13}=\frac{13}{13}=1$

Step2: Apply empirical rule for part (a)

The empirical rule states that for a normal distribution, approximately 68% of data lies within $z=\pm1$.

Step3: Find bounds for 99.7% (part b)

The empirical rule says 99.7% of data lies within $z=\pm3$. Calculate the bounds:
Lower bound: $\mu - 3\sigma = 47 - 3\times13 = 47 - 39 = 8$
Upper bound: $\mu + 3\sigma = 47 + 3\times13 = 47 + 39 = 86$

Answer:

(a) 68%
(b) 8 and 86