Suppose gre quantitative scores are normally distributed with a mean of 588588 and a standard deviation of 154154. a university plans to send letters of recognition to students whose scores are in the top 12%12%. what is the minimum score required for a letter of recognition? round your answer to the nearest whole number, if necessary.

The minimum score required for a letter of recognition is 769

Step-by-step explanation:

If GRE Quantitative scores are normally distributed with a mean of 588 and a standard deviation of 154, then, for a value x, the associated z-score is computed as (x-588)/154. On the other hand, we are looking for a value "a" such that P(X > a) = 0.12 or equivalently P(X < a) = 0.88. But this last expression is equivalent to P((X-588)/154 < (a-588)/154) = 0.88, i.e., (a-588)/154 is the 88th quantile of the standard normal distribution. Therefore (a-588)/154 = 1.1750, and so, a = 588 + (154)(1.1750) = 768.95

hey there!

12 is twice as much as 6 because 6 goes into 12 2 times.

since the rabbit is twice as tall, we should multiply the weight of the first rabbit, 0.25, by 2.

0.25 x 2 = 0.5

therefore, the 12 inch chocolate rabbit weighs 0.5 lb, or half a pound.

hope this !

answer

a the answer hopes this

step-by-step explanation: