According to the data, the mean quantitative score on a standardized test for female college-bound high school seniors was 500. The scores are approximately Normally distributed with a population standard deviation of 50. A scholarship committee wants to give awards to college-bound women who score at the 99th percentile or above on the test. What score does an applicant need? Include a well-labeled Normal curve as part of your answer.

Step-by-step explanation:

Hello!

The variable of interest is X: Score that a female college student obtained in a standardized test.

This variable has an approximately normal distribution with standard deviation σ= 50 and mean μ= 500

You need to find the score value that marks the 99th percentile, meaning that it separates 99% of the distribution from the top 1%, symbolically:

P(X≥x₀)= 0.01 or P(X≤x₀)= 0.99

Using the standard normal distribution Z=(X-μ)/σ ≈ N(0;1) you have to use the standard normal distribution to find the corresponding value.

first using the Z-table you have to look the Z value that accumulates 0.99 of distribution:

z₀= 2.334

Using the Z formula and the known values of μ and σ you can clear the value of x₀

z₀= (x₀-μ)/σ

z₀*σ= x₀-μ

x₀= (z₀*σ) + μ

x₀= (2.334*50)+500= 616.7

The college student has to become a score of 616.7 or more to receive the award.

I hope it helps!

what is the question here

step-by-step explanation:

25.2

step-by-step explanation:

2+4x6-8/10= 25.2