Question 469024: I need a tutor to review these and tell me if I am performing the operations correctly. I am not sure if I am going about it right...
Suppose the scores of certain standardized test are normally distributed with mean 500 and standard
deviation 60.
8. (1 point) What percent of the test takers have a score of 440 or less? Round answer to four places after
the decimal point.
500-440= 60/60 = 1
1= .8413
9. (1 point) What percent of the test takers have a score of 590 or more? Round answer to four places
after the decimal point.
590-500= 90/60= 1.5
1.5= .9332
10. (1 point) What percent of the test takers have a score between 440 and 590? Round answer to four
places after the decimal point.
.9332-.8413= 1.5 - 1 = .6915
11. (1 point) Suppose a college will only admit students who score in the top 20% among all test takers,
what will be the minimum required score be? Round answer to the nearest whole number.
.20= .8416 = x-500= 80.496= 580
Answer by ccs2011(207)   (Show Source):
You can put this solution on YOUR website!
Z = (x-u)/s, where u=mean, s=std deviation
The probabilities given always are for less than or equal to that number.
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8. you used 1 instead of -1:
Always subtract the mean, if its smaller you should get a negative Z
For Z = -1, P = .1587
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9. Look at question, they want scores greater than 590.
For Z= 1.5, P = .9332
So 93% are less, the compliment is 1-P
1-P = .0668
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10. Don't subtract Z values, only P values
Take P from 590 and subtract P from 440
.9332 - .1587 = .7745
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11. Top 20% also means that at that score 80% will be less than that score.
Find Z such that P=0.8
Corresponding Z is .8416
Find x using this value.
(x-500)/60 = .8416
x-500 = 50
x = 550
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