The time required for a citizen to complete the 2000 U.S. Census "Long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. what proportion of the citizens will require less than 1 hour?
97.72%
Use the tool at
http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html
Is there anyway you could explain that to me so I can understand it?
1 hour is 20 minutes = 2 standard deviations away from the mean. Your stats book should have a cumulative normal distribution table. Look up 2.0 in the table. I get .9772. Ergo, 97.72% will require less than an hour.
To find the proportion of citizens who will require less than 1 hour to complete the 2000 U.S. Census "Long" form, we need to find the area under the normal distribution curve for the given mean and standard deviation.
The first step is to convert the given 1 hour into minutes, which is 60 minutes. We want to find the proportion of citizens who will require less than 60 minutes, so we need to calculate the area to the left of that point on the normal distribution curve.
Using the properties of the normal distribution, we can calculate the z-score using the formula:
z = (x - μ) / σ
where x is the value we are interested in, μ is the mean, and σ is the standard deviation.
In this case, x = 60 minutes, μ = 40 minutes, and σ = 10 minutes.
z = (60 - 40) / 10
z = 2
Now that we have the z-score, we can use a standard normal distribution table or a calculator to find the proportion of citizens who will require less than 1 hour.
Looking up the z-score of 2 in a standard normal distribution table, we find that the area to the left of z = 2 is approximately 0.9772.
This means that approximately 97.72% of citizens will require less than 1 hour to complete the census form.
Alternatively, you can use a calculator or statistical software to calculate the cumulative probability directly. The cumulative probability for z = 2 is 0.9772, which gives us the same result.
Therefore, approximately 97.72% of citizens will require less than 1 hour to complete the 2000 U.S. Census "Long" form.