Download Solutions to Practice Exam 3

Solutions to Practice Exam 3
Math 203
March 16, 2012
1. Suppose that on any given day, the probability you will like one of the meals at the
“Explore” station in Anderson Commons is 53 . During next week, Sunday through
Saturday, what is the probability you will like an “Explore” meal on
(a) exactly 5 of those days?
Solution. This is a binomial trial question. We are asking for exactly 5 successes
out of 7 days, where the probability of success is 35 . So n = 7, k = 5, p = 53 , and
q = 25 , and the answer is
5 2
3
2
C(7, 5)
= 0.2612736.
5
5
(b) at least 5 of those days?
Solution. Now we want at least 5 successes; in other words, we want either 5, 6,
or 7 successes. So our answer is
5 2
6 1
7 0
3
3
3
2
2
2
C(7, 5)
+ C(7, 6)
+ C(7, 7)
= 0.419904
5
5
5
5
5
5
2. Let X be the random variable corresponding to how many pounds of candy a random
trick-or-treater in Moorhead gets on Halloween. Suppose the probability distribution
of X is given by the following table.
x
Pr(X = x)
0.5 1 1.5 2
0.3 0.4 0.2 0.1
(a) Find the expected value E(X), also known as µ.
Solution. For a probability distribution, the formula for µ is
µ = x1 p 1 + x2 p 2 + · · · + xn p n .
Thus the answer is
µ = (0.5)(0.3) + (1)(0.4) + (1.5)(0.2) + (2)(0.1) = 1.05.
(b) Find the standard deviation σ.
Solution. The formula for the standard deviation of a probability distribution is
p
σ = (x1 )2 p1 + (x2 )2 p2 + · · · + (xn )2 pn − µ2 ,
so the answer is
p
√
σ = (0.5)2 (0.3) + (1)2 (0.4) + (1.5)2 (0.2) + (2)2 (0.1) − 1.052 = 0.2225 ≈ 0.4717.
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Practice Exam 3 Solutions
Math 203, Spring 2012
3. Your friend Joe is a big film fan, and on average he likes 75% of the films he sees.
During the Fargo Film Festival, Joe skips his classes and abandons his homework to
watch 48 films in a row. Let the random variable X denote how many of those films
he will like.
(a) What is the formula giving the probability distribution of X? In other words,
what is Pr(X = k)?
Solution. This is a binomial distribution with n = 48, p = 0.75, and q = 1 − p =
0.25. So the probability distribution is
Pr(X = k) = C(48, k)(0.75)k (0.25)48−k .
(b) How many of the 48 movies do you expect that Joe will like?
Solution. For a binomial distribution, µ = E(X) = np = 48 · 0.75 = 36.
(c) What is the standard deviation of the random variable X?
√
√
√
Solution. For a binomial distribution, σ = npq = 48 · 0.75 · 0.25 = 9 = 3.
(d) Use the normal distribution to approximate the probability that Joe will like
at least 30 of the 48 movies.
Solution. We will pretend that X, instead of a binomial random variable, is
instead a normal random variable with µ = 36 and σ = 3. Since we want at least
30, the cutoff should be 29.5. So
29.5 − 36
Pr(X ≥ 30) = Pr(X ≥ 29.5) ≈ Pr Z ≥
= Pr(Z ≥ −2.1667)
3
= 1 − Pr(Z ≤ −2.1667)
≈ 1 − 0.0158
= 0.9842.
4. A publisher of textbooks has 1 misprint per 50 pages. The misprints occur randomly.
In 100 pages, what is the probability of
(a) no misprints?
Solution. Since the misprints occur randomly, this is a Poisson distribution. We
are looking at 100 pages, so
1 misprint
λ misprints
=
;
50 pages
100 pages
thus λ = 2. Using the formula Pr(X = k) =
λk −λ
e ,
k!
k Pr(X = k)
0
0.13534
1
0.27067
2
0.27067
..
..
.
.
Thus Pr(X = 0) = 0.13534.
we fill in the table
Practice Exam 3 Solutions
Page 3 of 4
Math 203, Spring 2012
(b) at most 2 misprints?
Solution. Using the table above, we see Pr(X ≤ 2) = Pr(X = 0) + Pr(X =
1) + Pr(X = 2) = 0.13534 + 0.27067 + 0.27067 = 0.67668.
(c) more than 2 misprints?
Solution. Pr(X > 2) = 1 − Pr(X ≤ 2) = 1 − 0.67668 = 0.32332.
5. Suppose the weights of pumpkins in Hornbacher’s are normally distributed with mean
µ = 6 lb. and standard deviation σ = 2 lb.
(a) What fraction of the pumpkins weigh at least 10 pounds?
Solution. Our random variable X is normally distributed with µ = 6 and σ = 2.
The fraction of pumpkins over 10 pounds equals the probability that a randomly
selected pumpkin weights more than 10 pounds, so the fraction we want is
10 − 6
= Pr(Z ≥ 2) = 1−Pr(Z ≤ 2) ≈ 1−0.9772 = 0.0228.
Pr(X ≥ 10) = Pr Z ≥
2
(b) What fraction of the pumpkins weigh between 4 and 8 pounds?
Solution. Again we use the table (which would be given at the back of the test)
to compute that
8−6
4−6
≤Z≤
= Pr(−1 ≤ Z ≤ 1) = 1 − 2 Pr(Z ≤ −1)
Pr(4 ≤ X ≤ 8) = Pr
2
2
≈ 1 − 2(0.1587)
= 0.6826.
6. The grade point average (GPA) of graduating seniors at Tippecanoe College is normally
distributed with a mean of 2.7 and a standard deviation of 0.5. The top 8% of the class
(by GPA) will graduate cum laude. What is the minimum GPA required to graduate
cum laude?
Solution. Let m stand for the minimum GPA. Then we know that 8% of the population earns a GPA greater than or equal to m; in other words,
0.08 = Pr(X ≥ m).
Applying our rules, then, we see that
0.08 = 1 − Pr(X ≤ m) = 1 − Pr(Z ≤
m − 2.7
)
0.5
so Pr(Z ≤ m−2.7
) = 0.92. Looking in the table, we find that Pr(Z ≤ 1.41) = 0.9207,
0.5
so we want
m − 2.7
= 1.41.
0.5
Solving for m, we find m = 2.7 + 0.5 · 1.41 = 3.405.
Practice Exam 3 Solutions
Page 4 of 4
Math 203, Spring 2012
7. If a cow gets a certain disease, there is only a 25% probability that it will recover. If
200 cows get the disease, what is the probability that at least 60 will recover? Give an
approximate answer using a normal approximation.
Solution. This is a binomial distribution, with n = 200, p = 0.25, and q = 0.75.
√
Thus the mean is µ = np = 200(0.25) = 50 and the standard deviation is σ = npq =
p
200(0.25)(0.75) ≈ 6.1237. Because we are approximating a binomial distribution
with a normal distribution, we need to adjust by 0.5. Thus
59.5 − 50
Pr(X ≥ 60) = Pr(X ≥ 59.5) ≈ Pr Z ≥
≈ Pr(Z ≥ 1.55).
6.1237
Then using the table, we look up and find
Pr(Z ≥ 1.55) = 1 − Pr(Z ≤ 1.55) = 1 − 0.9394 = 0.0606.