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A psychology professor assigns letter grades on a test according to the following scheme. A: Top 7% of scores B: Scores below the top 7% and above the bottom 64% C: Scores below the top 36% and above the bottom 25% D: Scores below the top 75% and above the bottom 6% F: Bottom 6% of scores Scores on the test are normally distributed with a mean of 78.4 and a standard deviation of 7.6. Find the minimum score required for an A grade. Round your answer to the nearest whole number, if necessary.

User Gks
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Answer:

The minimum score required for an A grade is 89.8.

Explanation:

We are given that a psychology professor assigns letter grades on a test according to the following scheme. A : Top 7% of scores. B : Scores below the top 7% and above the bottom 64%. C : Scores below the top 36% and above the bottom 25%. D : Scores below the top 75% and above the bottom 6%. F : Bottom 6% of scores.

Scores on the test are normally distributed with a mean of 78.4 and a standard deviation of 7.6.

Let X = Scores on the test

SO, X ~ Normal(
\mu=78.4,\sigma^(2) =7.6^(2))

The z-score probability distribution for normal distribution is given by;

Z =
(X-\mu)/(\sigma) ~ N(0,1)

where,
\mu = mean time = 78.4


\sigma = standard deviation = 7.6

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, the minimum score required for an A grade so that it represents Top 7% of scores is given by;

P(X
\geq x) = 0.07 {where x is the required minimum score

P(
(X-\mu)/(\sigma)
\geq
(x-78.4)/(7.6) ) = 0.07

P(Z
\geq
(x-78.4)/(7.6) ) = 0.07

So, the critical value of x in the z table which represents the top 7% of the area is given as 1.4996, that is;


(x-78.4)/(7.6) =1.4996


{x-78.4}{} =1.4996* 7.6


x = 78.4 + 11.39696 = 89.8 or 90

Hence, the minimum score required for an A grade is 89.8.

User Sommmen
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