The gre is an entrance exam that most students are required to take upon entering graduate school. in 2014, the combined scores for the verbal and quantitative sections were approximately normally distributed with a mean of 310 and a standard deviation of 12. what percent of scores were between 286 and 322? round your answer to the nearest whole number.

82%

Step-by-step explanation:

Data:

mean, μ = 310

standard deviation, σ = 12

We need to use the standard normal distribution table (see figures attached).

Z is calculated as follows:

Z = (x - μ)/σ

Replacing with x = 286 and x = 322:

Z = (286 - 310) / 12 = -2

Z = (322 - 310) / 12 = 1

So, we need to find:  

P(-2 ≤ Z ≤ 1) =

= P(-2 ≤ Z ≤ 0) + P(0 ≤ Z ≤ 1)

From the first table:

P(-2 ≤ Z ≤ 0) = 0.5 - P(Z ≤ -2) = 0.5 - 0.0228 = 0.4772

From the second table:

P(0 ≤ Z ≤ 1) = 0.3413

Then,

P(-2 ≤ Z ≤ 1) = 0.4772 + 0.3413 = 0.8185, which as a percentage is 0.8185*100 = 81.85% ≈ 82%

add all the numbers and check the ones place.

the answer is 486. as the ones place is more than 4, the estimation will be the next 10 ie, 490