Suppose that the height of Mc female students has normal distribution with mean

Question

Suppose that the height of Mc female students has normal distribution with mean 62 inches and standard deviation 8 inches. a. Find the height below which is the shortest 30% of the female students. b. Find the height above which is the tallest 5% of the female students 8.MENSA is an organization whose members possess IQs in the top 2% of the population. If IQs are normally distributed, with mean 100 and a standard deviation of 16, what is the minimum IQ required for admission to MENSA 9. Let X be a normal random variable with mean = 12 and standard deviation = 2. Find the X corresponding to the 10th percentile of this distribution 10. Entry to a certain University is determined by a national test. The scores on this test are normally distributed with a mean of 500 and a standard deviation of 100. Tom wants to be admitted to this university and he knows that he must score better than at least 70% of the students who took the test. Tom takes the test and scores 585. Will he be admitted to this university? 11. Let X be scores on a computer skills test with mean = 100 and standard deviation = 10. Assume the scores follow a normal distribution. Find the 90th percentile for these scores

Explanation / Answer

7)

for bottom 30%; zscore =-0.5244

hence corresponding height =mean +z*std deviation=62-0.5244*8=57.8048

for top 5%; z=1.645

hence corresponding height =mean +z*std deviation=62+1.645*8=75.1588

8)

for top 2%; z=2.0537

hence corresponding score =mean +z*std deviation=100+2.0537*16=132.86

9)

for 10-th percentile ; zscore =-1.2816

corresponding X =12-1.2816*2=9.4369

10)

for 70 percentile; z=0.5244

therefore corresponding score =500+0.5244*100=552.44

as Tom score is higher then that ; therefore he should get admission

11)

for 90th percentile; z=1.2816

therefore corresponding score =100+1.2816*10=112.816

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