Question

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.

Answer 1

Answer:

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:

$88\%=\frac{88}{100}=0.88$

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:

$z=\frac{x-\mu}{\sigma}$, where,

z = z-score,

x = Sample score,

$\mu$ = Mean,

$\sigma$ = Standard deviation.

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

$0.81057 =\frac{x-461}{124}$

Let us solve for x.

$0.81057\cdot 124=\frac{x-461}{124}\cdot 124$

$100.51068=x-461$

$100.51068+461=x-461+461$

$561.51068=x$

$x=561.51068\approx 562$

Therefore, the minimum score required for recruitment is 562.

Answer 2

Answer: it’s 606 I just got it wrong