Question

The Wall Street Journal (February 15,1972 ) reported that General Electric was sued in Texas for sex discrimination over a minimum height requirement of \(5 \mathrm{ft}\) 7 in. The suit claimed that this restriction eliminated more than \(94 \%\) of adult females from consideration. Let \(x\) represent the height of a randomly selected adult woman. Suppose that \(x\) is approximately normally distributed with mean 66 in. (5 ft 6 in.) and standard deviation 2 in. a. Is the claim that \(94 \%\) of all women are shorter than \(5 \mathrm{ft}\) 7 in. correct? b. What proportion of adult women would be excluded from employment as a result of the height restriction?

Short answer

The claim that 94% of all women are shorter than 5 ft. 7 in. is incorrect. Only approximately 69.15% of women are shorter than that. Also, approximately 69.15% of women would be excluded from employment due to the height restriction of 5 ft. 7 in.

Step-by-step solution

Analyze the Claim

First, you must calculate the Z-score to understand where a height of 5 ft. 7 in. stands within the given distribution. 5 ft. 7 in. is equal to 67 inches. The Z-score is defined by the formula: \(Z = \frac{x - \mu}{\sigma}\). Where \(x\) is the observation (67 inches in this case), \(\mu\) is the mean (66 inches), and \(\sigma\) is the standard deviation (2 inches). So the Z-score for 67 inches is \(Z = \frac{67 - 66}{2}\) which equals to 0.5.

Validate the Claim by finding the proportion

Now we need to verify the claim that more than 94% of adult women are shorter than 5 ft. 7 in. This can be done by finding the area to the left of the Z-score in a standard Normal Distribution. Area to the left denotes the proportion of the population that is less than the specific value. The value corresponding to Z=0.5 in standard Normal Distribution tables or using technology comes out to be 0.6915 or 69.15%. This shows that only about 69.15% women are shorter than 5 ft. 7 in. Thus the claim is incorrect.

Find the proportion of women excluded due to height restriction

The proportion of adult women that would be excluded from employment due to the height restriction is simply the area to the left of the Z-score (0.5) in a standard Normal Distribution, which we found to be 0.6915 or 69.15%. This means that approximately 69.15% of women would be excluded due to the height restriction of 5 ft. 7 in.