Airlines want to design seats so they are wide enough to fit 98% of all males. A

Question

Airlines want to design seats so they are wide enough to fit 98% of all males. A recent study found that men's hip widths follow a normal distribution with a mean of 14.4 in and a standard deviation of 1.01 in. 1. a. b. C. What percent of all men would fit in a seat that is 16 in wide? What percent of all men have hip widths between 13 and 17 in? How wide do the seats need to be to hold 98% of all men? 2. IQ scores are normally distributed with a mean of 100 and a standard deviation of 15 (Wechsler Test) a. Find the probability a randomly selected adult has an IQ less than 130. b. Find the probability that a randomly selected adult has an IQ greater than 131.5 c. Find the probability that a randomly selected adult has a score between 90 and 110. d. Find the probability that a randomly selected adult has an IQ between 110 and 120. e. Find the 10th percentile. f. Find the 60h percentile. g. Find the IQ score separating the top 35% from the rest.

Explanation / Answer

Mean is 14.4 and s is 1.01. z is given as (x-mean)/s

a) P(x<16) =P(z<(16-14.4)/1.01)=P(z<1.58), from normal distribution table we get 0.9429

b) P(13<x<17) =P((13-14.4)/1.01<z<(17-14.4)/1.01)=P(-1.39<z<2.57) or P(z<2.57)-P(z<-1.39) or P(z<2.57)-(1-P(z<1.39))= 0.9949-(1-0.9177)=0.9126

c) for 0.98, the z value can be found form normal distribution table as 2.06

thus answer is mean+s*z=14.4+2.06*1.01=16.4806

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