Question

Suppose the scores, x, on a college entrance exam are normally distributed with a mean of 500 and a standard deviation of 100. This college will only consider admission for those applicants whose scores exceeded the 86th percentile of the distribution. Find the minimum score an applicant must receive for admissions consideration.

Answer 1

Answer:

The minimum score an applicant must receive for admission is 392.

Step-by-step explanation:

Let the minimum score required be 'x₀'.

Given:

Mean score (μ) = 500

Standard deviation (σ) = 100

Percentage required for admission, P > 86% or 0.86

So, we are given the area under the normal distribution curve to the right of z-score which is 86%.

The z-score table gives the area left of the z-score value. So, we will find the z-score value for area 100 - 86 = 14% or 0.14

So, for value equal to 0.1401, the z-score = -1.08

Now, $P(x>x_0)=P(z>-1.08)$

So, we find x₀ using the formula of z-score which is given as:

$z=\frac{x_0-\mu}{\sigma}\\\\-1.08=\frac{x_0-500}{100}\\\\x_0-500=-1.08\times 100\\\\x_0=-108+500=392$

Therefore, the minimum score an applicant must receive for admission is 392.