Ergonomics. Exercises 9–16 involve applications to ergonomics,...

Question

Ergonomics. Exercises 9–16 involve applications to ergonomics, as described in the Chapter Problem.

Aircraft Cockpit The overhead panel in an aircraft cockpit typically includes controls for such features as landing lights, fuel booster pumps, and oxygen. It is important for pilots to be able to reach those overhead controls while sitting. Seated adult males have overhead grip reaches that are normally distributed with a mean of 51.6 in. and a standard deviation of 2.2 in.

a. If an aircraft is designed for pilots with an overhead grip reach of 53 in., what percentage of adult males would not be able to reach the overhead controls? Is that percentage too high?

Accepted Answer

Welcome back, everyone. In a basketball team, the vertical jump heights of players are normally distributed with a mean of 28 inches and a center deviation of 4 inches. To assess the player's fitness, the coach sets a benchmark vertical jump height of 32 inches. What percentage of players are expected not to meet this benchmark? A says 31.73%, B 84.13%, C 15.87%, and D 68.27%. So what we're going to do is simply draw a normal curve for this situation. Let's draw a bell-shaped curve. We are given our mean value. It says 28 inches, and specifically, we are also given our standard deviation. We will not label it on our graph, and we can basically say that the benchmark is at 32 inches. So that would be our random variable X, we're going to set it equal to 32, and now we have to understand what region we are solving for while we want to identify the area above X of 32, right? So we are solving for the probability of X being greater than 32. And we can express it as 1 minus the probability of X being less than 32, using the properties of probability, right? Essentially we can identify the region on the left and subtract that area from 1 to get the remaining area marked in red. So now what we are going to do is some standardized, and that would be the probability of X being greater than 32 is equal to 1 minus the probability of Now, let's express X in terms of Z. So, it's a standardized, we have to recall that Z equals x minus. The mean value divided by standard deviation. So, essentially we're going to say that this is equal to 1 minus P of Z being less than axis 32. Our new mean value is 28, and we're dividing by the standard deviation of 4. So this can be simplified into 1 minus the probability of C being less than 1. And then we can use the tables to identify the probability of C being less than 1. We get 1 minus 0.8413. And we simply want to subtract these two values. We get 0.1587. And in this percentage form that's 15.87%, we can simply multiply by 100 to show that the correct answer corresponds to the answer choice C. Thank you for watching.

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Key Concepts

Normal Distribution

Z-Score

Percentile

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