Seat Designs. In Exercises 7-9, assume that when seated, adult males have back-to-knee lengths that are normally distributed with a mean of 23.5 in. and a standard deviation of 1.1 in. (based on anthropometric survey data from Gordon, Churchill, et al.). These data are used often in the design of different seats, including aircraft seats, train seats, theater seats, and classroom seats. 7. Find the probability that a male has a back-to-knee length greater than 25.0 in. 8. Find the probability that a male has a back-to-knee length between 22.0 in. and 26.0 in. 9. Find the probability that nine males have back-to-knee lengths with a mean greater than 23.0 in.

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Seat Designs. In Exercises 7-9, assume that when seated, adult males have back-to-knee
lengths that are normally distributed with a mean of 23.5 in. and a standard deviation of
1.1 in. (based on anthropometric survey data from Gordon, Churchill, et al.). These data
are used often in the design of different seats, including aircraft seats, train seats, theater
seats, and classroom seats.
7. Find the probability that a male has a back-to-knee length greater than 25.0 in.
8. Find the probability that a male has a back-to-knee length between 22.0 in. and 26.0 in.
9. Find the probability that nine males have back-to-knee lengths with a mean greater than 23.0 in.
Transcribed Image Text:Seat Designs. In Exercises 7-9, assume that when seated, adult males have back-to-knee lengths that are normally distributed with a mean of 23.5 in. and a standard deviation of 1.1 in. (based on anthropometric survey data from Gordon, Churchill, et al.). These data are used often in the design of different seats, including aircraft seats, train seats, theater seats, and classroom seats. 7. Find the probability that a male has a back-to-knee length greater than 25.0 in. 8. Find the probability that a male has a back-to-knee length between 22.0 in. and 26.0 in. 9. Find the probability that nine males have back-to-knee lengths with a mean greater than 23.0 in.
Expert Solution
Step 1

From the provided information,

Mean (µ) = 23.5

Standard deviation (σ) = 1.1

X~N (23.5, 1.1)

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