Verified Answer :

LammettHash

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

\(\mathbb P(X

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

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

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

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