Practice Quiz with Solutions: Chapter 6 (20 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.

Solution:  This is a binomial distribution: each student passes or not, and whether one student passes has nothing to do with whether anyone else passes.

μ = np = 6*0.8 ⇒ μ = 4.8 people

σ = √(npq) = √(6*0.8*0.2) ⇒ σ = 0.9798

A coin is weighted — the chance of heads is not 50%. On five flips of that coin, the probability of various numbers of heads is shown by this probability distribution:

Answer:  discrete — Every probability distribution is either discrete or continuous, and this one asks a “how many” question.

Common mistake: Quite a few students answer “continuous”. Ask yourself: when flipping a coin 5 times, could you get 2.3 heads? No, only certain specified values are allowed, namely 0 to 5 in whole numbers; therefore this must be a discrete PD.

Solution: 

        x’s in L1, P(x)’s in L2
        1-VarStats L1, L2

The answers can be written down from the output screen: μ = 2.0000, σ = 1.0954

Common mistake: Remember that in a probability distribution you are talking about a population: the correct symbols are μ and σ, not x̄ and s.

Remark:  Remember the interpretation: if you flipped the coin in many, many sets of five flips, and wrote down the number of heads in each set, the average would be 2.0000 heads per set and the standard deviation would be 1.0954.

Common mistake: You must specify both lists in the 1-Var Stats command when you have a frequency distribution or a probability distribution.

Common mistake: Some students try to use μ=np and σ=√(npq). But p, the probability of a head on one flip, was not given, so you can’t use these formulas. For problems like this one, where all you have is a set of x’s and P(x)’s, you can’t use the formulas but must follow the procedure on page 228.

Solution:  This is a binomial distribution: each person is either helped or not, and the probability is the same from one person to the next. n = 5, p = 0.4, and x = 3, 4, or 5.

BINOMPRB program: n=5, p=.4, from=3, to=5

answer: 0.31744

Common mistake: “At least three” means three or more (3, 4, or 5), not more than three (4 or 5).

Alternative solution: If you don’t have the program, you can use the TI-83 or TI-84 native function binompdf:

binompdf(5, .4) → L3

L3(4)+L3(5)+L3(6) = 0.3174

If you do it that way, remember that the row numbers in L3 are 1 more than the values of x. If you entered L3(3)+L3(4)+L3(5) you got the incorrect answer of 0.6528. If your solution was otherwise correct, you got half credit.

Answer: x = 0, 1, 2, 3, 4, or 5

Solution:  This is a binomial distribution: a person is either a smoker or not, and the probability is 55% for each random person. n = 4, p = 0.55, x = 2.

BINOMPRB program

N=4, P=0.55, from=2, to=2

Answer: 0.3675

Alternative solution:  binompdf(4, .55, 2) = 0.3675