The grades on the last math exam had a mean of 88%. Assume the population of grades on math exams is known to be normally distributed with a standard deviation of 6. What percent of students earn a score between 72% and 90%?

0.6255

0.5843

0.6754

0.5346

Question

The grades on the last math exam had a mean of 88%. Assume the population of grades on math exams is known to be normally distributed with a standard deviation of 6. What percent of students earn a score between 72% and 90%?

 0.6255

 0.5843

 0.6754

 0.5346

.whitedragon. · Answer

@ganeshie8

anonymous · Answer

@mane1

anonymous · Answer

yes

.whitedragon. · Answer

type something brooklyn ill give medal 2 u

kirbykirby · Answer

You need to find $P(72<X<90)$. You must standardize your random variable.
$$P(72<X<90)=P\left( \frac{72-88}{6}<\frac{X-88}{6}<\frac{90-88}{6}\right)=P(-2<Z<\frac{1}{3})$$.

anonymous · Answer

Okay thank you! I just didn't understand how to standardize it with two numbers. So it's the same as with one?

anonymous · Answer

now I just find the z score right?

kirbykirby · Answer

oops $P\left( \frac{-8}{3}<Z<\frac{1}{3}\right)$

kirbykirby · Answer

i divided by 8 instead f 6 by accident

anonymous · Answer

It's fine!

kirbykirby · Answer

So yeah you look at the z-scores of -8/3 = -2.66, and z=1/3=0.33  on a normal table. And determine what the probability is between those 2 values.

anonymous · Answer

help mane1

anonymous · Answer

urgent

anonymous · Answer

Wait so how do I do that? I am actually an intelligent person I just missed a day of class and my professor didn't explain it too well to me.

kirbykirby · Answer

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