Answer :
The score that separates the top 59% from the bottom 41% is approximately 39.1 .
To find the score that separates the top 59% from the bottom 41% in a normally distributed set of scores, you can use the z-score formula:
[tex]\[ z = \frac{{X - \mu}}{{\sigma}} \][/tex]
Where:
- [tex]\( X \)[/tex]is the score you want to find,
-[tex]\( \mu \)[/tex] is the mean (37.3),
- [tex]\( \sigma \)[/tex]is the standard deviation (8).
First, we need to find the z-score corresponding to the 59th percentile. We can find this using a standard normal distribution table or a calculator. The z-score corresponding to the 59th percentile is approximately 0.261.
So, we can set up the equation:
[tex]\[ 0.261 = \frac{{X - 37.3}}{{8}} \][/tex]
Now, solve for [tex]\( X \)[/tex]
[tex]\[ X - 37.3 = 0.261 \times 8 \]\[ X - 37.3 = 2.088 \]\[ X = 2.088 + 37.3 \]\[ X = 39.388 \][/tex]
Therefore, the score that separates the top 59% from the bottom 41% is approximately 39.4.
The closest option provided is:
b. 39.1
