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 5%?

Answer 1

The top 5% of scores belong to the 95th percentile, which means the cutoff score $k$ is such that

$\mathbb P(X<k)=0.95$

Transforming to the standard normal distribution, you have

$\mathbb P(X<k)=\mathbb P\left(\dfrac{X-76.4}{6.1}<\dfrac{k-76.4}{6.1}\right)=\mathbb P(Z<k^*)=0.95$

A left-tail probability of 95% corresponds to a z-score of about $k^*=1.6449$, which means the cutoff score must be around

$1.6449=\dfrac{k-76.4}{6.1}\implies k\approx86.4$

Answer 2

Using the normal distribution, it is found that the minimum score to be in the top 5% is of 86.4.

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In a normal distribution, the z-score is used.

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

• $\mu$ is the mean and $\sigma$ is the standard deviation.

• It measures how many standard deviations the measure is from the mean.
• Each z-score has an associated p-value, which is the percentile of the measure X.

In this problem:

• Mean of 76.4, thus $\mu = 76.4$
• Standard deviation of 6.1, thus $\sigma = 6.1$
• The top 5% is at least the 100 - 5 = 95th percentile, which is X when Z has a p-value of 0.95, so X when Z = 1.645.

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

$1.645 = \frac{X - 76.4}{6.1}$

$X - 76.4 = 6.1(1.645)$

$X = 86.4$

The minimum score needed to be in the top 5% is of 86.4.

A similar problem is given at https://brainly.com/question/15289375