Question

Suppose that scores on a particular test are normally distributed with a mean of 140 and a standard deviation of 16. What is the minimum score needed to be in the top 20% of the scores on the test? Carry your intermediate computations to at least four decimal places, and round your answer to one decimal place.

Answer 1

Solution:

Let X be a random variable which represents the score on a particular test.

Given that, X ~ N(140, 16²)

i.e.  μ = 140 and σ = 16

If a score is in top 20% it means that at least 80% of the scores is less than that score. Hence, to obtain the minimum score needed to be in the top 20% of the scores on the test, we need to obtain 80th percentile of the random variable X. The value of the 80th percentile will be the minimum score that is needed.

Let 80th percentile be x_{​​​​​​1}.

We know that if X ~ N(μ, σ^{​​​​​​2}) then

.................................(1)

Using Z-table we get, P(Z < 0.8416) = 0.80

Comparing P(Z < 0.8416) = 0.80 and equation (1) we get,

  

The 80th percentile is 153.4656.

Hence, the minimum score needed to be in the top 20% of the scores on the test is 153.4656.

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