Question

Engineers want to design seat in commercial aircraft so that they are wide enough to fit 95% of all males ( Accommodating 100% of males would require very wide seats that would be much too expensive)Men ha e hip breadth that are normally distributed with a mean of 14.2 in and a standard deviation of 1.1in find P 95 that is , find the hip breadth for men that separates the smallest 95% from the largest 5%

Answer 1

To find the hip breadth for men that separates the smallest 95% from the largest 5%, we need to find the 95th percentile value of a normal distribution with a mean (\(\mu\)) of 14.2 inches and a standard deviation (\(\sigma\)) of 1.1 inches.

Steps:

1. Identify the z-score that corresponds to the 95th percentile from the standard normal distribution.
2. Use the z-score to find the corresponding hip breadth using the formula for converting a z-score to an x value.

Step 1: Find the z-score for the 95th percentile

From standard normal distribution tables or a calculator, the z-score that corresponds to the 95th percentile is approximately \(z = 1.645\).

Step 2: Convert the z-score to hip breadth

Use the z-score formula:

\[ x = \mu + z \cdot \sigma \]

Now plug in the values:

\[ x = 14.2 + (1.645 \cdot 1.1) \]

Step 3: Perform the calculation

\[ x = 14.2 + (1.645 \cdot 1.1) \approx 14.2 + 1.8095 \approx 16.0095 \]

Final Result

The hip breadth that separates the smallest 95% from the largest 5% is approximately 16.01 inches. This means that a seat designed to accommodate 95% of males would need to be at least 16.01 inches wide.