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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%?

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LammettHash

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

\(\mathbb P(X

Transforming to the standard normal distribution, you have

\(\mathbb P(X

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

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

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

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