How Common Are Millionaires Among Ford Employees?

[Changoo]

I hope someone can tell me where I am going wrong with this dang homework problem. There are several questions to this one problem, but once I figure this out, I will be able to figure the other parts of this question out.


Approximately 5% of US Families have a net worth in excess of 1 million. A survery in the year 2000 found that 30 percent of Ford's employees were millionaires. If random samples of 100 Ford's employees had been taken that year what proportion of the samples would have been between 25% and 35%?

Here I go:

100(.30)=30
100(1-.30)=70

thus, .25-.30/(square root of) (.30)(.70)/100=-.05/.0458=-1.092

Z=.1379

and

.35-.30/.0458=1.092

Z=.8621


I am lost here. Can someone help??

 

[wais]

It looks like you are on the right track with your calculations. However, to find the proportion of samples between 25% and 35%, you need to find the area under the normal curve between the two z-scores you calculated (Z=0.1379 and Z=0.8621). You can use a standard normal table or a calculator to find this area, which corresponds to the proportion of samples between 25% and 35%.

Alternatively, you can also use the formula for the standard normal distribution to find this proportion. The formula is P(a < Z < b) = Φ(b) - Φ(a), where Φ(x) is the cumulative standard normal distribution function. In this case, a = 0.1379 and b = 0.8621.

P(0.1379 < Z < 0.8621) = Φ(0.8621) - Φ(0.1379) = 0.8051 - 0.5578 = 0.2473

Therefore, approximately 24.73% of the samples would fall between 25% and 35%. I hope this helps!

 

[M98Ranger]

It seems like you have the right idea, but may have made a few calculation errors. Let's break down the problem step by step to see where you may have gone wrong.

First, we know that 30% of Ford's employees were millionaires in the year 2000. This means that out of 100 employees, 30 would have a net worth of at least 1 million.

Next, we need to find the proportion of samples that would fall between 25% and 35%. To do this, we need to calculate the z-score for each of these percentages.

For 25%, we have:

Z = (.25 - .30) / √(.30)(.70)/100 = (-.05) / .0458 = -1.092

For 35%, we have:

Z = (.35 - .30) / √(.30)(.70)/100 = .05 / .0458 = 1.092

Now, to find the proportion of samples between these two z-scores, we need to use a z-table. Looking at the table, we can see that the area between -1.092 and 1.092 is approximately 0.6764.

This means that approximately 67.64% of the samples would fall between 25% and 35%.

I hope this helps clarify the problem for you. Remember, when working with proportions and z-scores, it's important to double check your calculations and use a z-table to find the proportion. Good luck with the rest of your homework!