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 percentage of scores that fell between 58.9 and 75.1?(1 point) Responses 47.5% 47.5% 95% 95% 81.5% 81.5% 68%

Answer 1

To find the percentage of scores that fall between 58.9 and 75.1 on a standardized test with a normal distribution, we can use the properties of the normal distribution.

1. **Calculate the z-scores** for 58.9 and 75.1 using the formula:

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

where \( \mu \) is the mean and \( \sigma \) is the standard deviation.

- For \( X = 58.9 \):

\[
z_1 = \frac{(58.9 - 64.3)}{5.4} \approx \frac{-5.4}{5.4} = -1
\]

- For \( X = 75.1 \):

\[
z_2 = \frac{(75.1 - 64.3)}{5.4} \approx \frac{10.8}{5.4} = 2
\]

2. **Use the z-scores to find the corresponding percentiles**:

- A z-score of -1 corresponds approximately to the 15.87th percentile.
- A z-score of 2 corresponds approximately to the 97.72th percentile.

3. **Calculate the percentage of scores between these two z-scores**:

\[
P(58.9 < X < 75.1) = P(Z < 2) - P(Z < -1) \approx 0.9772 - 0.1587 = 0.8185
\]

4. **Convert to percentage**:

\[
0.8185 \times 100 \approx 81.85\%
\]

Thus, the best approximation of the percentage of scores that fell between 58.9 and 75.1 is approximately **81.5%**.