An engineering professor wants to design motorcycle helmets. He found that men have head breadths that are normally distributed with a mean of 6.0 in. and standard deviation of 1.0 in. based on anthropometric survey data from Gordon, Churchill, et al. (a) If a man is randomly selected, find the probability that his head breadth is less

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An engineering professor wants to design motorcycle helmets. He found that men have
head breadths that are normally distributed with a mean of 6.0 in. and standard deviation
of 1.0 in. based on anthropometric survey data from Gordon, Churchill, et al.
(a) If a man is randomly selected, find the probability that his head breadth is less
than 6.2 in.
(b) A new helmet company plans an initial production run of 100 helmets. Find the
probability that 100 randomly selected men have a mean head breadth is less than
6.2 in.
(c) The product manager sees the result from part (b) and reasons that all helmets
should be made for men with head breadths less than 6.2 in., because they would fit
all but few men. But the professor does not agree that reasoning. Provide the
professor's argument.
Transcribed Image Text:An engineering professor wants to design motorcycle helmets. He found that men have head breadths that are normally distributed with a mean of 6.0 in. and standard deviation of 1.0 in. based on anthropometric survey data from Gordon, Churchill, et al. (a) If a man is randomly selected, find the probability that his head breadth is less than 6.2 in. (b) A new helmet company plans an initial production run of 100 helmets. Find the probability that 100 randomly selected men have a mean head breadth is less than 6.2 in. (c) The product manager sees the result from part (b) and reasons that all helmets should be made for men with head breadths less than 6.2 in., because they would fit all but few men. But the professor does not agree that reasoning. Provide the professor's argument.
Expert Solution
Step 1

(a)

Consider the probability that his head breadth is less than 6.2 in is,

P(X<6.2)=P(X-μσ<6.2-μσ)                =P(z<6.2-6.01.0)               =P(z<0.2)               =0.5793 From the Excel function, =NORM.DIST(0.2,0,1,TRUE)

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