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 add quite a bit of 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.

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

z_J = (1390−1000)/200 = 390/200 = 1.95

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

Because he has the higher z score, according to the test Jacinto is more intelligent.

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

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

Solution: You need the class marks, which are the midpoints of the classes, as shown at right. They 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, 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: Remember that 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) x̄ = 499.81 (before rounding, 499.8051948)

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

(d) s² = 162.38 (before rounding, 162.3801036)

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

Common mistake: σx on the screen is not the variance. The TI-83 doesn’t automatically compute the variance, but you must compute it. See Descriptive Statistics of a Data Set on the TI-83/84.

Common mistake: You must not square the rounded value 12.74. If you do, you get 162.31. See The Big No-no.

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, 36924, 149380, 169541

For the box plot, see Descriptive Statistics of a Data Set on the TI-83/84.

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