Question

Scores on an exam follow an approximately normal distribution with a mean of 76.4 and a standard deviation of 6.1 points. what is the minimum score you would need to be in the top 10%?

Answer 1

The 90th percentile is the upper 10% of a distribution with a cutoff score $x$ such that

$\mathbb P(X<x)=\mathbb P\left(\dfrac{X-76.4}{6.1}<\dfrac{x-76.4}{6.1}\right)=\mathbb P(Z<z)=0.90$

where $z$ is the corresponding z-score for $x$.

In the standard normal distribution, this cutoff value is about $z\approx1.2816$, so in terms of $x$ this is

$1.2816=\dfrac{x-76.4}{6.1}\implies x\approx84.2$

Answer 2

From the normal distribution, the minimum score needed to be in the top 10% is of 84.2.

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Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean, and has a p-value associated, which is the percentile of the measure X.

In this problem:

• Mean of 76.4 means that $\mu = 76.4$
• Standard deviation of 6.1 means that $\sigma = 6.1$.
• The top 10% of scores are at least the 100 - 10 = 90th percentile, which is X when Z has a p-value of 0.9, so X when Z = 1.28.

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 76.4}{6.1}$

$X - 76.4 = 1.28(6.1)$

$X = 84.2$

The minimum score you would need to be in the top 10% is of 84.2.

A similar problem is given at https://brainly.com/question/2934460