The combined math and verbal scores for students taking a national standardized examination for college admission, is normally distributed with a mean of 560 and a standard deviation of 260. If a college requires a student to be in the top 35 % of students taking this test, what is the minimum score that such a student can obtain and still qualify for admission at the college?
(round to the nearest integer

The minimum score that such a student can obtain and still qualify for admission at the college = 660.1

Step-by-step explanation:

This is a normal distribution problem, for the combined math and verbal scores for students taking a national standardized examination for college admission, the

Mean = μ = 560

Standard deviation = σ = 260

A college requires a student to be in the top 35 % of students taking this test, what is the minimum score that such a student can obtain and still qualify for admission at the college?

Let the minimum score that such a student can obtain and still qualify for admission at the college be x' and its z-score be z'.

P(x > x') = P(z > z') = 35% = 0.35

P(z > z') = 1 - P(z ≤ z') = 0.35

P(z ≤ z') = 1 - 0.35 = 0.65

Using the normal distribution table,

z' = 0.385

we then convert this z-score back to a combined math and verbal scores.

The z-score for any value is the value minus the mean then divided by the standard deviation.

z' = (x' - μ)/σ

0.385 = (x' - 560)/260

x' = (0.385×260) + 560 = 660.1

Hope this Helps!!!

step-by-step explanation:

6.48/4 4 goes into 6.48 1.62 times