M203E Practice Quiz 5

Instructions: The quiz will be open book (any books) and open notes (any notes or written material). 

[1] (20 points) Consider the data 7, 8, 8, 1, 5, 6, 6, 9, 11, 20. 
(a) Find the mean of this data: (7 + 8 + 8 + 1 + 5 + 6 + 6 + 9 + 11 + 20)/10 = 81/10 = 8.1
(b) Find the median of this data. First put the data in order: 1 5 6 6 7 8 8 9 11 20. 
Thus the median is (& + 8)/2 = 7.5.
(c) Find the first quartile for this data. Find the median of the bottom half of the data:
1 5   6   6 7 So the first quartile is 6.
(d) Find the third quartile for this data. Find the median of the top half of the data:
8 8   9   11 20 So the first quartile is 9.
(e) Find the standard deviation of this data. 
Find the average of the squares of the deviations: 
[(7-8.1)^2+(8-8.1)^2+(8-8.1)^2+(1-8.1)^2+(5-8.1)^2+(6-8.1)^2+(6-8.1)^2+
(9-8.1)^2+(11-8.1)^2+(20-8.1)^2]/10 = 220.9/10 = 22.09.
Now take the square root of this: (22.09)^(1/2) = 4.7 is the standard deviation.

[2] (5 points) Give a set of numbers such that the mean is bigger than the median. Also 
compute the mean and median for the numbers you give.

Consider 1, 2, 6. The mean is (1+2+6)/3 = 9/3 = 3, but the median is 2. Note that 6 is 
an outlier which pulls the mean up.

[3] (5 points) Suppose you take a poll of Lincoln residents by selecting a random 
sample of phone numbers from the phone book. Discuss ways that this sample may be biased. 

Not everyone has a phone, so this sample is biased against non-phone users. 
Also, phone books just give numbers for landlines, so this sample is biased 
against people who have only a cell phone. Probably folks who have only a 
cellphone and no landline tend to be younger so this sample may well be 
biased toward older people.

[4] (10 points) IQ's are normally distributed with mean 100 and standard 
deviation 15. Recall that 68% of a normally distributed population is within 
1 standard deviation of the mean, 95% is within 2 standard deviations of the 
mean, and 99.7% is within 3 standard deviations of the mean. 

(a) What percentage of the population has an IQ above 130? Show how you obtain your answer.

Since 130 is 2 standard deviations above the mean, of the 5% not within 
2 standard deviations of the mean, 2.5% will be more than 2 standard deviations 
above the mean with the other 2.5% being more than 2 standard deviations below 
the mean. Thus the answer is: 2.5% have IQ's above 130.

(b) What percentage of the population has an IQ in the range 70 to 115? 
Show how you obtain your answer. 

Since 70 is 2 standard deviations below the mean, of the 95% within 
2 standard deviations of the mean, 47.5% = 95%/2 will be between 70 and 100. 
Likewise, 68%/2 = 34% will be between 100 and 115 (since 115 is 1 standard deviation 
above the mean). Thus the answer is: 47.5 + 34 = 81.5% of the population has 
an IQ in the range 70 to 115.

[5] (10 points) Assume a newspaper polls a random sample of 400 of the 350000 
adults living in Omaha and finds 80% of the Omahans sampled are satisfied with 
their water quality, with a margin of error of ± 4%.

(a) What is the chance that the actual percentage of all 350000 adults in 
Omaha who are satisfied with their water quality is not in the range from 76% to 84%?

The margin of error typically is for a 95% level of confidence, so 5% 
will be outside the range 76% to 84%.

(b) Suppose the newspaper polled 400 people randomly selected from all 
250 million adults living in the U.S., and that 80% percent of these 400 
were satisfied with their water quality. Which poll gives a more reliable 
measurement? I.e., is the 80% estimate for Omaha's water quality 
satisfaction rate likely to be more accurate, less accurate or equally 
as accurate as the 80% estimate for the water quality satisfaction rate 
for the whole U.S.? Which poll has a larger margin of error? Explain your answer.

Both estimates have the same reliability because both used random samples of the same size.