TC3 → Stan Brown → Statistics → Chapter 6 Practice Quiz
revised 17 Oct 2012

# 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 often add some extra commentary. Please see Show Your Work for the what, why, and how.

1(points: 2½)  Suppose 80% of students who register for Elizabethan Sonnets complete the course successfully. And suppose you took many, many samples of six people, with replacement. What would be the expected mean and standard deviation of the number of people that would finish successfully, per sample of six?

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 = √[np(1−p)] = √(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:
 x P(x) 0 1 2 3 4 5 0.0778 0.2591 0.3456 0.2305 0.0768 0.0102
2(points: 1) What type of probability distribution is this? Answer continuous, discrete, binomial, or categorical.

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.

Remark: A case could be made for answering “binomial”, although the distribution is not presented that way. There are only two outcomes, heads and tails, and obviously the probability of a head is the same from one flip to the next. Give yourself credit if you answered “binomial” for that specific reason. The problem is that you don’t know that probability, so you can’t use binomial probability calculations and you have to fall back to the generic discrete probability distribution.

3(points: 2½) Find the mean and standard deviation of the above probability distribution, and label them with their correct symbols.

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 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=√[np(1−p)]. 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 303 of Sullivan, Michael, Fundamentals of Statistics 3/e (Pearson Prentice Hall, 2011).

4(points: 2)  Long experience shows that a particular drug will help 40% of the people who take it. If you take a random sample of five people, what is the probability that the drug helps at least three?

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.

MATH200A part 3: n=5, p=.4, from=3, to=5

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.

5(points: 1)  You roll five dice and count the number of twos that appear. List the possible values of the discrete random variable, X = “number of twos in five dice”.

Answer: x = 0, 1, 2, 3, 4, or 5

6(points: 2)  Suppose 55% of the people in the area are smokers. If you take a random sample of four people, what is the probability that less than two of them are smokers?

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 = 0 or 1.

MATH200A part 3

n=4, p=0.55, from=0, to=1

Alternative solution: binomcdf(4, .55, 1) = 0.2415

7(points: 4)  A lottery has a 1 in 10 million chance of paying \$10,000,000, a 1 in 125 chance of paying \$100, and a 1 in 20 chance of paying \$10. A ticket costs \$5, and you do not get that money back if you win a prize.
(a) Construct a discrete probability distribution.
(b) Is this a good deal or a bad deal for you? Explain.
x (in L1)P(x) (in L2)
\$9,999,9951/10,000,000
951/125
51/20
−50.9419999

Solution: (a) Since you lose the \$5 ticket price even if you collect a prize, you need to subtract \$5 from each prize to get your net winnings. If you don’t win any prize, your net is −5. The first three probabilities are given, and the last one is found by subtracting the others from 1. The result is shown at right.

(b) To find whether it’s a good deal, compute the mean or expected value. A positive expected value means a good deal for you; a negative expected value means a bad deal for you. Use 1-VarStats L1,L2 to obtain μ = \$−2.70. This is a bad deal for you because on average you lose \$2.70 per ticket.

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