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: it’s 606 I just got it wrong

Answer 2

562.

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\%=(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=(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 =(x-461)/(124)`

Let us solve for x.

`0.81057\cdot 124=(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.