Question

On a standardized test with a normal distribution, the mean was 64.3 and the standard deviation was 5.4. What is the best approximation of the percent of scores that fell between 58.9 and 75.1?

81.5%
95%
47.5%
68%

Answer 1

To find the percentage of scores that fall between 58.9 and 75.1 in a normal distribution with a mean of 64.3 and a standard deviation of 5.4, we need to calculate the z-scores corresponding to these values.

The z-score is calculated using the formula:

\[
z = \frac{(X - \mu)}{\sigma}
\]

where:
- \(X\) is the value,
- \(\mu\) is the mean, and
- \(\sigma\) is the standard deviation.

1. Calculate the z-score for 58.9:

\[
z_{58.9} = \frac{(58.9 - 64.3)}{5.4} = \frac{-5.4}{5.4} = -1
\]

2. Calculate the z-score for 75.1:

\[
z_{75.1} = \frac{(75.1 - 64.3)}{5.4} = \frac{10.8}{5.4} = 2
\]

Now we have two z-scores: \(z_{58.9} = -1\) and \(z_{75.1} = 2\).

Next, we will use the standard normal distribution (z-table) to find the percentage of data between these two z-scores.

- The area to the left of \(z = -1\) (from z-table) is approximately 0.1587 (or 15.87%).
- The area to the left of \(z = 2\) (from z-table) is approximately 0.9772 (or 97.72%).

Now, to find the percentage of scores between 58.9 and 75.1:

\[
\text{Percentage} = P(Z < 2) - P(Z < -1) = 0.9772 - 0.1587 \approx 0.8185
\]

So, \(0.8185\) corresponds to approximately \(81.85\%\).

Thus, the best approximation of the percent of scores that fell between 58.9 and 75.1 is:

**81.5%**.