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Normally Distributed

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Question 1. The graduate selection committee wants to select the top 10% of applicants. On a standardized test with a mean of 500 and a standard deviation of 100. What would be the cutoff score for selecting the 10% of applicants, assuming that the standardized test is normally distributed?
Mean = 500; deviation = 100; x = 10%
(100-10)/100=.90 (looking at the normal distribution curve, the closest I could get was .8997 which rounds up to 0.90) and the Z score is 1.28.
1.28*100 = 128 and the mean is 500 so I will add it to the 128 (128+500=628).
So the cutoff point would be 628.

Question2
The average commute time via train from the Chicago O’Hare Airport to downtown is 60 minutes with a standard deviation of 15 minutes. Assume that the commute times are normally distributed. What proportion of commutes would be: a. Longer than 80 minutes (80-60)/15 = 1.333 (z > 1.333)
1.33 = .092 (9.2%) b. Less than 50 minutes
(50-60)/15 = -.667 (z< -.667)
-.667 = .251 (25.1%) c. Between 45 and 75 minutes
(45-60)/15 = -1.000
-1.00 = .1587 = .159
(75-60)/15 = 1.000
1.000= .8413 = .841
.841 - .159 = 0.682*100 = 68.2%

Question 3
Bob takes an online IQ test and finds that his IQ according to the test is 134. Assuming that the mean IQ is 100, the standard deviation is 15 and the distribution of IQ scores is normal, what proportion of the population would score higher than Bob? Lower than Bob?

(134 – 100)/15 = z 34/15 = 2.2666
Z = 2.27 = .0116 (1.16%)
Bob is a very smart guy with a 2.27 z score which mean very few people (1.16%)has a higher IQ. Then there would be 98.84% who has a lower IQ than Bob. 1.16% higher than Bob
98.84% lower than

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