Question 1189614: in an inteligient test administered to 1000 students the average score was 42 and the standared deviation 24. find (a) the number of students exceeding a score of 50 (b) the number of students lying between 30 and 54 (c) the value of score exceeded the top 100 students.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! answers to your questions are shown below:
mean = 42
standard deviation = 24.
n = 1000
z-score = (x - m) / s
z is the z-score.
x is the raw score
m is the mean
s is the standard deviation, in this case.
using the z-score table at https://www.rit.edu/academicsuccesscenter/sites/rit.edu.academicsuccesscenter/files/documents/math-handouts/Standard%20Normal%20Distribution%20Table.pdf, you would do the following:
find:
(a) the number of students exceeding a score of 50.
z = (x - m) / x becomes z = (50 - 42) / 24 = 8/24 = .33333.....
round to 2 decimal places to get z = 0.33
look in the z-score table for a z-score of 0.33 to get the area to the left of that = .62930.
that's the area to the left of a z-score of .33.
subtract that from 1 to get .3707.
the area to the right of a z-score of .33 is equal to .3707.
if you want the score to be more than, you are looking for the area to the right.
if you want the score to be less than, you are looking for the area to the left.
the z-score table referenced always gives you the area to the left.
to find the area to the right, subtract that from 1 as i did above.
.3707 is the area under the normal distribution curve to the right of a z-score of .33.
the number of students with a score greater than 50 would be .3707 * 1000 = 370.7 = 371 when you round to the nearest integer.
(b) the number of students lying between 30 and 54.
you need to find two z-scores.
first for the raw score of 30 and next for the raw score of 54.
you than need to get the area to the left of each.
you then need to subtract the smaller area from the larger area to get the area in between.
the lower z-score is z = (30 - 42) / 24 = -.5
the higher z-score is z = (54 - 42) / 24 = +.5
area to the left of a z-score of -.5 = .30854
area to the left of a z-score of +.5 = .69146
area in between = .69146 minus .30854 = .38292.
number of students with a score between 30 and 54 is .38292 * 1000 = 382.92 = 383.
(c) the value of score exceeded the top 100 students.
the top 100 students will consume 10% of the total number of students.
you are looking for the area to the right of a z-score that is closest to .10.
since the table only gives you the area to the left of a z-score, then you are looking for the area to the left of a z-score that is closest to 1 - .10 = .9.
in the z-score table, you get the following:
z-score of 1.28 has an area of .89973 to the left of it.
z-score of 1.29 has an area of .90147 to the left of it.
since .89973 is closer to .9 than .90147, you would choose a z-score o 1.28, unless you want to do a manual interpolation or use a z-score calculator.
the z-score of 1.28 in the z-score formula gives you the following:
1.28 = (x - 42) / 24
solve for x to get:
x = 1.28 * 24 + 42 = 72.72.
the score that has 10% of the scores above it would be 73 when rounded to the nearest integer.
let me know if you have any questions.
theo
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