SOLUTION: Having a difficult time with this problem...Please help! Solve the problem. A standardized math test was given to 6000 students. The scores were normally distributed with a m

Question 1164995: Having a difficult time with this problem...Please help!
Solve the problem.
A standardized math test was given to 6000 students. The scores were normally distributed with a mean of 380 and a standard deviation of 50. How many students scored between 380 and 480?

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
test was given to 6000 students.
mean was 380.
standard deviation was 50.
how many students scored between 380 and 480.

doing it the easy way.
use the calculator found at:

http://davidmlane.com/hyperstat/z_table.html

enter the mean of 380.
enter the standard deviation of 50.
choose between.
enter 380 to 480.

the calculator tells you that .4772 of the area under the normal distribution curve is between 380 and 480

multiply 6000 by that ratio to get .4772 * 6000 = 2863.2.
round this to 2863.

that should be your answer.

the hard way is a little different but should give you the same answer, or something close to it.

your mean is 380 and your standard deviation is 50.
you want to find out how many students scored between 380 and 480.
use the z-score table to find the z-score for 380 students and the z-score for 480 students.

the z-score formula is z = (x - m) / s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation.

the mean is 380.
the standard deviation is 50.

for a raw score of 380, the formula becomes z = (380 - 380) / 50 = 0.

for a raw score of 480, the formula becomes z = (480 - 380) / 50 = 2.

to to the z-score table and look for the area to the left of a z-score of 0 and to the left of a z-score of 2.

i used the table that can be found at:

https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf

from this table, i found that:

the area to the left of a z-score of 0 is equal to .5.
the area to the left of a z-score of 2 is equal to .97725.

the area between those two z-scores is equal to .97725 minus .5 = .47725.

this is actually a little more detailed than the answer given by the online calculator.
this is because the online calculator does some rounding or truncating of the number of digits.
because of that, the table is a little more accurate.
but, as i said before, the answer will be close.

you now do the same thing you did before.
.47725 * 6000 = 2863.5 which you would round off to 2863 or to 2864 according to your preference.
remember:
you got 2863.2 before and now you got 2863.5.
they're pretty close and more than good enough for a rough estimate.

i used the z-score calculator in the TI-84 Plus and i got the following:
m3an = 380
standard deviation = 50
portion of the normal distribution curve between 380 and 480 is equal to .4772499375.
that * 6000 = 2863.499625.

this is again, quite close to 2863.2 and to 2863.5.

so, take your pick.
you should state how you got the answer and then give the number of students you got by using that method.
if you round up, then the answer should be 2864.
if you round down, then the answer should be 2863.

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