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SOLUTIONS ACTIVITY SET 7
Activity 7.1 In each part, indicate, (1) whether the variable is discrete or continuous AND (2) whether it
is binomial or not AND (3) if it is binomial, give values for n and p.
a. Number of times a “head” is flipped in 10 flips of a coin
1. Discrete
2. Binomial
3. n = 10, p = 0.5
b. Time to complete a 30 question multiple choice test
1. Continuous 2. Not Binomial since continuous
c. Number of correct answers on a 30 question multiple choice test for somebody who randomly guesses
at every question. There are four answer choices for each question,
1. Discrete
2. Binomial – counting correct answers 3. n = 30, p = 0.25
d. A woman buys a lottery ticket every week for which the probability of winning anything at all is 1/10.
She continues to buy them until she has won 3 times. X = the number of tickets she buys.
1. Discrete
2. Not binomial since the number of trials is not fixed
Activity 7.2 In Stat 200 last year students were asked to rate how much they liked various kinds of music
on a scale of 1 (don’t like at all) to 6 (like very much). Following is a probability distribution (with one
probability not given) for females’ rating of Top 40 music.
X=Rating
Probability
1
.04
2
.05
3
4
5
6
.09
??
.32
.26
a. Is X = music rating a discrete variable or a continuous variable? Explain.
Discrete. There are a small number of distinct possible outcomes (the ratings 1-6).
b. What must be the value of the probability for X = 4 (the probability that rating equals 4)? Explain how
you determined this.
P(x=4) is 0.24 since all the probabilities need to add to 1.
c. Determine E(X), the mean rating given by females. {HINT: Find (rating*probability) for each rating,
and then add up those values.}
E(X) = (1)(.04) + 2(.05) + 3(.09) + 4(.24) + 5(.32) + 6(.26) = .04+.10+.27+.96+1.60+1.56 = 4.53
d. Find the probability that the rating given is 3 or less. Note: When we find the probability that a variable
is less than or equal to some value it is called a cumulative probability. To find it, we add probabilities for
all values up to and including that point.
P(X≤3) = P(X=1)+P(X=2)+P(X=3) = 0.04 + 0.05 +0.09 = 0.18
e. For each rating value, determine the cumulative probability.
X = Rating
Cumulative
Probability
1
2
3
4
5
6
.04
.09
.18
.42
.74
1.0
f. Use the cumulative probabilities just found as an aid in finding the probability that the rating of a
randomly selected student would be greater than 4. Show work.
P (rating > 4) = 1 – P(rating ≤ 4) = 1- .42 = .58
Activity 7.3 A roulette wheel has 38 numbers. A player can bet on one specific number. If that number
comes up when the wheel is spun, the player gains $35. If the number does not come up, the player loses
$1. Let X = the amount won or lost in a single play of this game.
a. Give values for the probability distribution of X. You can express the probabilities as fractions.
X = Amount
+35
−1
1/38
37/38
won or lost
Probability
b. Find E(X), the mean value of X. Show work.
E(X) = (35)(1/38) +(-1)(37/38) = −0.053
c. Write sentence that interprets the value of E(X) in this situation, for somebody with no training in
statistics. Over many plays (or, in the long run) a player will lose an average of 5.3 cents per play.
Activity 7.4 Suppose somebody randomly guesses at every one of 20 True-False questions.
a. The number of correct guesses is a binomial random variable. What are the values of the parameters n
and p?
n = 20
p = 0.5
b. What is the expected number and standard deviation of correct guesses at the n = 20 answers? Show
how you determined this. Since this is a binomial experiment we can use the following formulas:
Expected value: np = 20(0.5) = 10
Standard Deviation:
np(1  p) =
20*0.5 p(1  0.5) = 2.24
Use Minitab to find the probability for parts c through f. Instructions on how to do this are given in
the Week 6 folder of the course website.
c. Assuming random guessing, what is the probability that the number of correct guesses is 13 or fewer?
0.942341
d. What is the probability that somebody just guessing could get 14 or more correct? Show work (Hint:
This is the complement of the previous part.)
1-0.942341 = 0.057659
e. What is the probability that somebody just guessing would get 18 or more correct? (Hint: You’ll have
to deal with the probability of 17 or less first.
1 − 0.999799 = 0.000201
f. What is the probability that somebody just guessing gets exactly 10 correct? (You’ll have to click on
Probability rather than Cumulative Probability in the Minitab dialog box.)
0.176197
g. What is the probability that the number correct for somebody just guessing is one of 9, 10, or 11
correct? Hint: First find separate probabilities for each of 9, 10, and 11. Show any work.
P(9 right) + P(10 right) + P(11 right) =
0.160179 + 0.176197 + 0.160179 = 0.496555