SOLUTION: In a Statistics examination, the average score was 80 and the variance was 9. Students with scores ranging from 83 to 88 received a grade of 1.50. If the scores are approximately n

Question 1173215: In a Statistics examination, the average score was 80 and the variance was 9. Students with scores ranging from 83 to 88 received a grade of 1.50. If the scores are approximately normally distributed and 11 students received a grade of 1.50, determine the percentage of students with scores lower than 83.
Let X be the scores in Statistics exam.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
mean = 80
standard deviation = square root of variance = sqrt(9) = 3

z-score is found by the following formula:
z = (x - m) / s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation

when x = 83, z = (83 - 80) / 3 = 3 / 3 = 1
when x = 88, z = (88 - 80) / 3 = 8 / 3 = 8/3
area to the left of a z-score of 1 is equal to .841345
area to the left of a z-score of 8/3 is equal to .996170
area in between 83 and 88 is equal to .996170 - .841345 = .154825

11 students had grades between 83 and 88.
11/.154825 = approximately 71 students in total.
.841345 * 71 = approximately 60 students who got grades less than 83.
the balance of the students got grades greater than 88 = 1 - .84345 - .154825 = .00382 * 71 = less than 1.
rounded that becomes 0.

the numbers don't really come out very well because they are not whole numbers and the percentages, especially of the scores above 88, don't relate very well to a whole number.
in fact, a percentage that low would lead to a whole number of 0.

to answer the question, i think all you had to do was find the area under the normal distribution curve to the left of a score of 83.
the rest of the information was superfluous, since it wasn't necessary to answer the problem.

to answer the problem, all you had to do was find the area under the normal distribution curve to the left of a score of 83.
the z-score was (83 - 80) / 3 = 1
the area to the left of a z-score of 1 was .841345.
that's 84.1345%.
that would be your answer, without any further analysis.
it was not necessary to determine the number of students, like i attempted to do.
in fact, i'm kind of sorry i did go the extra mile.

i would recommend going with 84.1345% rounded to whatever number of decimal points required.

let me know how you did with that.
theo

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