Suppose that an airline uses a seat width of 17.1 in. Assume men have hip breadths that are normally distributed with a mean of 14.9 in. and a standard deviation of 0.9 in. Complete parts (a) through (c) below. (a) Find the probability that if an individual man is randomly selected, his hip breadth will be greater than 17.1 in. The probability is-
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![### Probability of Seat Width Accommodation
Suppose that an airline uses a seat width of 17.1 inches. Assume men have hip breadths that are normally distributed with a mean of 14.9 inches and a standard deviation of 0.9 inches. Complete parts (a) through (c) below.
#### (a) Find the probability
Find the probability that an individual man is randomly selected, his hip breadth will be greater than 17.1 inches.
**The probability is:**
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In this problem, you are tasked with determining the probability that a randomly selected man's hip breadth will exceed the seat width provided by the airline. This problem is an example of a question involving the normal distribution, specifically needing the calculation of the cumulative probability for values beyond a certain point.
To solve for the probability:
1. **Define the variables**:
- Mean ($\mu$): 14.9 inches
- Standard deviation ($\sigma$): 0.9 inches
- Value of interest (X): 17.1 inches
2. **Convert to a Z-Score**:
- The Z-score formula is given by \( Z = \frac{X - \mu}{\sigma} \)
- Plugging in the values: \( Z = \frac{17.1 - 14.9}{0.9} \)
3. **Find the Cumulative Probability**:
- Use standard normal distribution tables or software to find the probability corresponding to the Z-score calculated.
Remember that the result will give the probability of a man's hip breadth being less than or equal to 17.1 inches, so you will need to subtract this value from 1 to find the probability of being greater than 17.1 inches.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fab0e5d97-7b6b-46cc-bf12-d986af136b0a%2F1f0bd7e5-18ca-4c8e-a3b0-521f5edaf54c%2F6elx6to6_processed.jpeg&w=3840&q=75)
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