TC3 → Stan Brown → Statistics → Chapter 3 Practice Quiz
revised 12 Sep 2012

# Practice Quiz with Solutions: Chapter 3 (35 min)

Before looking at these solutions, please work the practice quiz.

Disclaimer: This quiz is representative of the level of difficulty you should expect, but it doesn’t include every single topic from the week’s work. The real quiz may include some other topics and may skip some that are in this practice quiz. (The real quiz also may not word questions in the same way as the practice quiz. You should focus on the concepts, not a particular form of words.)

See also: How to Take a Math Test

These solutions show about the same level of work I expect from you, though I often add some extra commentary. Please see Show Your Work for the what, why, and how.

1(points: 2½)  SAT verbal scores are normally distributed, with a mean of 500 and standard deviation of 100. You randomly select a test taker. What’s the probability that s/he scored between 500 and 700?

Solution: Since 500 equals the mean, its z score is 0. For 700, compute the z score as z = (700−500)/100 = 2. So you need P(0 < z < 2). Make a sketch and shade this area.

Draw an auxiliary line at z = −2. You know that the area between z = −2 and z = +2 is 95%, so the area between z = 0 and z = 2 is half that, 47.5% or 0.475.

See also: Empirical Rule Practice Problems has more problems of this type.

Remark: The diagram and calculation are sufficient for showing your work; you don’t need to write anything else. I included the above commentary for those who might have trouble finding the solution.

2(points: 2½) Mensa, the largest high-IQ society, accepts SAT scores as indicating intelligence. Assume that the mean combined SAT score is 1500, with standard deviation 300. Jacinto scored a combined 2070.

Maria took a traditional IQ test and scored 129. On that test, the mean is 100 and the standard deviation is 15.

From the test scores, who is more intelligent? Explain.

Solution: To compare apples and oranges, compute their z scores:

zJ = (2070−1500)/300 = 570/300 = 1.90

zM = (129−100)/15 = 29/15 = about 1.93

Because she has the higher z score, according to the tests Maria is more intelligent.

Remark: The difference is very slight. We might wonder whether on another day Jacinto might do slightly better and Maria slightly worse, reversing their ranking.

3(points: 1) In a particular data set, the mean is 8700 and the median is 5000. What if anything can you say about the shape of the distribution?

Answer: The mean is much greater than the median. This usually means that the distribution is skewed right, like incomes at a corporation.

Test Scores Frequencies, f
(L2)
Class Midpoints, x
(L1)
470.0–479.915475.0
480.0–489.922485.0
490.0–499.929495.0
500.0–509.950505.0
510.0–519.938515.0

4(points: 5)  At right is a sample shown as a grouped frequency distribution. Compute the following quantities and label each with its proper symbol: (a) sample size, (b) mean, (c) standard deviation. Round to two decimal places.

Use any valid method, but show your work. (Begin by filling in the third column including column heading.)

Solution: You start with the class marks or midpoints, as shown at right.

Common mistake: Class midpoints are halfway between successive lower bounds: (470+480)/2 = 475. It’s not correct to calculate them between lower and upper bounds, (470+479.9)/2=474.95.

The class midpoints go in a list, such as L1, and the frequencies go in another list, such as L2. (Either label the columns with the lists you use, as I did here, or state them explicitly: “class marks in L1, frequencies in L2”.) Now use the command `1-VarStats L1,L2` — you need to write down the command that you used.

Common mistake: You must specify both lists when finding statistics of a frequency distribution. Always look at the sample size first, and if it’s too small you probably forgot the second list.

(a) n = 154

(b) = 499.81 (before rounding, 499.8051948)

(c) s = 12.74 (before rounding, 12.74284519)

Common mistake: The standard deviation is 12.74 (Sx), not 12.70 (σ), because this is a sample and not the population.

5(points: 4) Here are the sale prices of 15 randomly selected houses in Joliet IL in December 2004, from Sullivan, Michael, Fundamentals of Statistics 2/e (Pearson Prentice Hall, 2008), page 171, problem 4:

138,820   169,541   135,512   149,143   140,794   153,146   99,000   136,924
136,833   115,000   124,757   128,429   157,216   149,380   136,529

Compute the five-number summary and make a box plot. (You might want to do that on your TI-83 and copy the plot to your paper.)

Solution: Put the numbers in a statistics list, such as L1, and write down what you did: “prices in L1”. Then execute this command and write it down:

1-VarStats L1

Scroll down and you can read off the five-number summary: 99000, 128429, 136924, 149380, 169541

For the box plot, see MATH200A Program part 2.

You can see from the plot that there are no outliers.

Remark: Your calculator can create the boxplot, but you have to label it yourself. There are two ways to label a boxplot. One way, shown here, places each number of the five-number summary under its point in the boxplot. The other way shows an x axis with a consistent scale, such as 75, 100, 125, 150, 175. Which way is best depends on your audience and what you’re trying to display.

This page is used in instruction at Tompkins Cortland Community College in Dryden, New York; it’s not an official statement of the College. Please visit www.tc3.edu/instruct/sbrown/ to report errors or ask to copy it.

For updates and new info, go to http://www.tc3.edu/instruct/sbrown/stat/