High School

A normal distribution has a mean of [tex]\mu = 80[/tex] with [tex]\sigma = 20[/tex]. What score separates the highest 15% of the distribution from the rest of the scores?

A. [tex]X = 95[/tex]
B. [tex]X = 92[/tex]
C. [tex]X = 100.8[/tex]
D. [tex]X = 65[/tex]

Answer :

Final answer:

The score that separates the highest 15% from the rest in a normal distribution with a mean of 80 and standard deviation of 20 is 100.8, option C.

Explanation:

The question is asking to find the score that separates the top 15% from the rest in a normal distribution. To solve this, we look up the z-score that corresponds to the 85th percentile (100% - 15% = 85%) of a normal distribution, which is approximately 1.04. We then use the formula X = μ + (z)(σ), where X is the desired score, μ is the mean, and σ is the standard deviation. Plugging in the given mean (μ = 80) and standard deviation (σ = 20), we get X = 80 + (1.04)(20) which equals 100.8.

Final answer:

The score that separates the highest 15% is approximately 72.3.

None of the given options is correct

Explanation:

To find the score that separates the highest 15% of the distribution from the rest, we can use the concept of z-scores and the standard normal distribution.

First, let's find the z-score corresponding to the highest 15% of the distribution. Since the normal distribution is symmetric, we know that 15% of the distribution lies above the mean (50% in total), leaving 35% below the mean.

Using a z-table or a calculator, we find that the z-score corresponding to 35% below the mean is approximately -0.385.

Next, we can use the z-score formula to find the corresponding score in the original distribution. The z-score formula is:

z = (X - μ) / σ

Rearranging the formula, we have:

X = μ + z * σ

Plugging in the values, we get:

X = 80 + (-0.385) * 20

X ≈ 80 - 7.7

X ≈ 72.3

Therefore, the score that separates the highest 15% of the distribution from the rest is approximately 72.3.

None of the answer choices provided match the correct score of approximately 72.3.