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Scores on an English test are normally distributed with a mean of 37.3 and a standard deviation of 8. Find the score that separates the top 59% from the bottom 41%.

A. 32.6
B. 39.1
C. 35.5
D. 42.0

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

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