Suppose GRE Verbal scores are normally distributed with a mean of 461461 and a standard deviation of 124124. A university plans to recruit students whose scores are in the top 12%12%. What is the minimum score required for recruitment? Round your answer to the nearest whole number, if necessary.

562.

Step-by-step explanation:

We have been given that GRE Verbal scores are normally distributed with a mean of 461 and a standard deviation of 124. A university plans to recruit students whose scores are in the top 12%. We are asked to find the minimum score required for recruitment.

Top 12% means whose score are 88% or above.

Let us convert 88% into decimal as:

Now we will find z-score corresponding to 0.88 using normal distribution table.

Z-score corresponding to 0.88 is 0.81057.

Now we will use z-score formula and solve for sample score as:

, where,

z = z-score,

x = Sample score,

= Mean,

= Standard deviation.

Upon substituting our given information in z-score formula, we will get:

Let us solve for x.

Therefore, the minimum score required for recruitment is 562.

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