ATH Omework Olutions: Due Friday, Nov 1st

M ATH 20 – H OMEWORK 6 S OLUTIONS !
due Friday, Nov 1st
Instructions: This assignment is due at the beginning of class. Please write the questions in the correct order.
If I cannot read your handwriting, you won’t receive full credit.
You may use Wolfram Alpha to compute any necessary sums or integrals. If you have trouble with this, let me know.
If you’re using facts about distributions to answer the questions, be very clear about which distribution
you’re using to model that problem and why that distribution is appropriate.
1. A school wishes to accept 2000 students for their freshman class, and they expect 20,000 applications.
In order to make the admissions decisions very easy, the only criterion they will use is SAT score. So,
their goal is to accept a student if and only if their SAT score is in the top 10%. However, because
their computer system is so old, the applications only come in one at a time, and they must decide
whether to accept or reject before moving on to the next application. Assuming that SAT scores
are normally distributed with a mean of 1000 and a standard deviation of 200, how should they set
the score threshold to end up with as close to 2000 students as possible? Give your answer first
symbolically (in terms of a pdf, cdf, etc), then use a normal distribution table1 to provide a numerical
answer.
Solution: We want to find the SAT score x such that 90% of scores are below x and 10% of scores are
above x. A symbolic answer can expressed in many different ways. Here are a few (make sure you
see why they’re all equivalent):
(a) Let F ( x ) be the cumulative distribution function for the normal distribution with mean 1000 and
standard deviation 200. Then, the x we want is the solution to the equation F ( x ) = 0.9.
(b) Let f ( x ) be the density function for the normal distribution with mean 1000 and standard devi-
ation 200. Then, the x we want is the solution to
󰁝 x
1 ( x −1000)2
0.9 = f ( x ) = √ e− 80000 .
−∞ 200 2π
(c) Let f ( x ) be the density function for the standard normal distribution with mean 0 and standard
deviation 1. Then, the answer we want is 1000 + 200x, where x is the solution to
󰁝 x
1 2
0.9 = f ( x ) = √ e− x /2 .
−∞ 2π
The last characterization is the easiest to relate to the normal distribution table. We look through the
table for the value closest to 0.90, and find that in a standard normal distribution, P( X ≤ 1.28) =
0.8997 and P( X ≤ 1.29) = 0.9015. We’ll use the first value because its probability is closer to 0.9.
Hence, the cutoff should be placed 1.28 standard deviations above the mean. This is
1000 + 200(1.28) = 1256 points.

1 For example: http://z-table.com. The numbers in the leftmost column represent the number of standard deviations from the
mean up to 1 decimal place, and the numbers along to top row then refine to the second decimal place. Please ask me if you have
questions about this.
1
2
2. The density function f ( x ) of a continuous random variable X is given by

󰀝
A + Bx2 , if x ∈ [0, 2]
f (x) = .
0, otherwise
If E [ X ] = 1/2, find A and B.

Solution: There are two unknowns, A and B, so we should look for two pieces of information. The
first is that E [ X ] = 1/2. The second is that f ( x ) is actually a density, telling us that the integral of
f ( x ) is 1. So,
󰁝 ∞
1= f ( x )dx
−∞
󰁝 2
= ( A + Bx2 )dx
0
󰀗 󰀘
B 3 2
= Ax + x
3 0
8
= 2A + B.
3
Since, E [ X ] = 1/2, we know
󰁝 ∞
1
= x f ( x )dx
2 −∞
󰁝 2
= x ( A + Bx2 )dx
0
󰀗 󰀘
A 2 B 4 2
= x + x
2 4 0
= 2A + 4B.
Solve the simultaneous equations 2A + (8/3) B = 1 and 2A + 4B = 1/2, we find A = 1 and B = −3/8.
Hence the density function is
󰀫
8
1 − x2 , if x ∈ [0, 2]
f (x) = 3 .
0, otherwise
3. Since 1851, exactly 116 hurricanes have hit Florida (this includes the years 1851 and 2016, but not
2017—only direct hits by hurricanes are counted, not tropical storms). In 2005, Florida was hit by four
hurricanes: Cindy, Dennis, Katrina, and Wilma. If the probability of hurricane strikes has remained
the same since 1851, what is the probability of Florida being struck by four or more hurricanes in the
same year?
Solution: This is a classic Poisson distribution. We’ve assumed that the probability of hurricane
strikes has remained the same2 , hence our rate is 116 hurricanes per 2016 − 1851 + 1 = 166 years. As
the question asks about a year time frame, we have to adjust the rate:
116/166 hurricanes
λ= .
1 year
2 In reality, a bad assumption.
3
Now, the probability of four or more hurricanes in the same year is
3 3
λk e−λ
1− ∑ P( X = k) = 1 − ∑ k!
≈ 0.005719 = 0.57%.
k =0 k =0
In other words, this is a “1 in 200 years” type of event.
4. In the solutions manual to a Calculus textbook, there is about one faulty solution per fifty questions.
In a book with ten chapters, each with one hundred questions, what is the probability that there are
at least 15 faulty solutions in the whole book? Give your answer two ways: first with a binomial
distribution, then with a Poisson approximation. Use Wolfram Alpha or some other tools to find both
answers numerically, and compare them.
Solution: This is exactly a binomial distribution, and approximately a Poisson distribution.
A “success” is a faulty solution, hence p = 1/50 = 0.02. There are 1000 total problems, and so the
probability that at least 15 are faulty is
1000 󰀕 󰀖󰀕 󰀖k 󰀕 󰀖1000−k
1000 1 49
P( X ≥ 15) = ∑ k 50 50
≈ 0.89747 = 89.7%.
k =15
Using a Poisson approximation, the rate (which needs to have “per book” as its unit measurement) is
20 faculty solutions
λ= .
1 book
Hence,
1000
λk e−λ
P( X ≥ 15) = ∑ k!
= 0.89513 = 89.5%.
k=15

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M ATH 20 – H OMEWORK 6 S OLUTIONS !

due Friday, Nov 1st

1000 + 200(1.28) = 1256 points.

2. The density function f ( x ) of a continuous random variable X is given by

If E [ X ] = 1/2, find A and B.

Now, the probability of four or more hurricanes in the same year is

In other words, this is a “1 in 200 years” type of event.

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