Law of Averages

Topics: Arithmetic mean, Normal distribution, Standard deviation Pages: 6 (1479 words) Published: September 23, 2012
13.People in a large population average 60 inches tall. You will take a random sample and will be given a dollar for each person in your sample who is over 65 inches tall. For example if you sample 100 people and 20 turn out to be over 65 inches tall, you get $20. Which is better: a sample of size 100 or a sample of size 1,000? Choose one and explain. Does the law of averages relate to the answer you give? In this case a sample size of 100 would be better. This can be explained using law of averages and also by looking at the formula for Margin of Error. ME = 2 * SD of population/SQRT(size of sample)

In this case, the SD of population would remain the same. The only factor or value changing is the sample size. Sample size of 100 would give a bigger ME value as compared to sample size = 1000. With higher value of ME the possibility of values for height over 65 is more. However, it does not completely confirm that more values would be over 65 for sure. Also looking at the law of averages for a sample mean, “with a large randomly selected sample, the sample average tends to be close to the true population average”. In this case it is 60 inches. “The larger the sample, closer is sample average to population average”. So if we take sample size of 1000, the average of sample would be closer to 60 inches. Here we are assuming that the sample size of 100 and 1000 is selected without any kind of bias. 18.Based on a simple random sample of one hundred an analyst estimates the average hourly wage earned by workers in a city to be $30 and computes the margin of error to be $5. Can we conclude from this that most workers there earn between $25 and $35 an hour? Is this the right interpretation for the margin of error? Margin of error as per the text is “it estimates the largest distance you would reasonably expect to see between sample average and population average”1. Based on the data provided above, with ME = $5 we can be confident that in the population from which the analyst took the sample, average workers earn in the range of $25 to $35. However, it does not mean that most workers earn between $25 and $35. That would not be the right interpretation for the margin of error.

23.Polls showed the two main candidates in the 2004 presidential election were nearly tied on the day before the election. To predict the winner a newspaper would like to have a poll that has margin of error of less than 1%. Roughly how large a sample would be needed for such a poll?

Using the formula2:

n = sample size
p = true percentage
ME = Margin of error
=> n = [{2 * sqrt(50 * 50)}/1]2
=> n = (2 * 50)2
=> n = 10,000
We need a sample size of atleast 10,000 people for the margin of error to be less than 1%

6.A large population of overdue bills has balances that follow a normal curve. When we take a sample of 100 of these the average is $500 and the SD is $100.

a)What statement can you make about the range $300 to $700? 95% of all overdue bills have balances in the $300 to $ 700. This can be explained in few ways:

SD =$100
Average = $500
Range = $300 to $ 700

If we take actual score of $300 and plug it in the following formula:

Standard units = (actual score – average)/SD
= (300 – 500)/100
= -2

For actual score = 700, we will get
Standard units = (700 – 500)/100
= 2
This means that the overdue bill balances are in the range of -2 to +2 standard units which is same as 95% of area under the normal curve.

Another way to explain this is much simpler – SD = $100.
(-2*100) = -200
(+2*100) = 200

Going +- two times SD from mean we get the range $300 and $700. This is the range in which 95% of all overdue bills have balances.

b)What statement can you make about the range $480 to $520? 95% confidence interval for the average balance of overdue accounts is from $480 to $520. If we calculate SE for avg using the following formula:

SE for avg = SD/sqrt(100) = 100/10 = 10
i.e. using the...
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