# Answers to Home work assignment 1 Farid Alizadeh Faculty of Management

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Answers to Home work assignment 1
Faculty of Management
Rutgers University
October 4, 1999
Problem1: Page 259 of text no. 7.16 (30 points)
(a 15 points) The random variables of interest here are
X=The number of occurrences of a particular of face (say 5).
Y=The amount of money we win (or lose)
X can be either 0, or 1 or 2, or 3. Y can be either -1, 0, 1, or 2. (alternative answer: The question is a bit vaguely formed so some of you
may have values of -1, 1,2,3 for Y, which will be accepted if you state your
assumptions clearly.) The RV X is should be immediately recognized as
having a binomial distribution with n = 3 and p = 1/5. Thus we get the
following probability density function for X and Y:
Event
X
Y
no face
0
-1
one face
1
0
two faces
2
1
three faces
3
2
Probability density function
of X and Y
3
1 0 5 3
= 125
0
6
6
216
3
1 1 5 2
= 25
1
6
6
1 2 5 1 72
3
5
= 72
2
6
6
3
0
3
1
5
1
= 216
3
6
6
Probability distribution function
of X and Y
125
216
125
25
200
216 + 72 = 216
125
25
5
215
216 + 72 + 72 = 216
125
25
5
1
216 + 72 + 72 + 216 =
1
(b 10 points) The RV we are interested in is Y the amount of winnings. This
is obtained by plugging into the following formula:
X
Expected value of winnings =
x × Pr[X = x] =
x
= −1 ×
=−
1
1
2
125
25
5
1
+0×
+1×
+2×
216
72
72
216
October 4, 1999
(c 1 point) Since the expected value of gain for the play in one play is -0.5 we
expect to lose 50 cents each time we play; the more we play the more our
expected losses.
(d 3 points ) The house wins 50 cents each time it plays, thus the more people
play the more its expected winnings.
(e 1 point) Unless you are willing to buy a certain bridge connecting Brooklyn
to Manhattan, you shouldn’t play this game to make money.
Problem2: 20 points
You should recognize again that the problem best fits binomial distribution with
“success” being the same as “prefers chocolate ice cream”. Then p = 0.70 and
n = 75. Now we need to know what is the probability that between 35 and 46
kids prefer chocolate among 75 kids. We observe that this event is the same as
Event 1: “fewer than 46 students prefer chocolate”
and
Event 2: “fewer than 36 kids prefer chocolate”
and out of these events we make:
Event 3: “Event 1 and not Event 2”
Thus we need to subtract the probability of the second event from the probability of the first one:
46 X
75
(0.70)i (0.3)75−i
i
i=0
35 X
75
Pr[Event 2] =
(0.70)i (0.3)75−i
i
Pr[Event 1] =
i=0
Pr[Event 3] = Pr[Event 1] − Pr[Event 2] =
46 X
75
(0.70)i (0.3)75−i
i
i=35
Now this is an ugly sum, you could go through and calculate it term by term,
but why? You can find the answer by using a computer or a calculator. For
instance using Excel you would enter the following formula in a cell:
BINOMDIST(46,75,0.7,TRUE)- BINOMDIST(75, 0.70, 35, TRUE)
which gives 0.0676 − 2.104 × 10−5 = 0.067573.
2
October 4, 1999
Problem3: Page 293 of text no. 8.16 (20 points)
Here we follow the model that the expected proportion is equal to probability.
Thus, assuming N(x; µ, σ) is the cumulative distribution function with mean µ
and standard deviation σ we get:
(a) We need to find √
Pr[T ≤ 6] = N(6; 12, 3) Pr[T ≤ 9] = N(9; 12, 3), since
µ = 12 and σ = 9 = 3. Using Excel you have to write in a cell
=NORMDIST(6,12,3,TRUE)
which equals 0.02275, and
=NORMDIST(9,12,3,TRUE)
which is 0.158655. Alternatively we can look up the table at the back of
the book, but first we need to standardize: We know that in general if
X has normal distribution with mean µ and variance σ then the random
variable Z = X−µ
has normal distribution with mean 0 and σ = 1. Thus
σ
we need to look up the standard normal table under
• To get the proportion for people getting returns within 6 weeks look
up Z = (6 − 12)/3 = −2 and we see 0.4772 but this the probability
of Z falling between 0 and 2. We need to subtract this number from
0.5 to get 0.5 − 0.4772 = 0.0228 which is the same as what we got
from Excel.
• same procedure as above: (9 − 12)/3 = −1 which corresponds to
0.3413. subtracting this from 0.5 gives: 0.5 − 0.3413 = 0.1587.
(b) We need to computer Pr[T > 15] = 1 − Pr[T ≥ 15] = 1 − N(15; 12, 3). In
Excel we need to enter
=1 - NORMDIST(15, 12, 9, TRUE)
to get 0.15866. A more clever way would be to observe that Pr[T ≥ 15] =
Pr[T ≥ 9] because they are on the opposite side of the the mean which is
12 with equal distance. Solving by looking up table is the same procedure
as we did in part (a) for 9.
(c) We need to find the value of t such that Pr[T ≤ t] = 0.9, that is we need to
find a point at which N(t; 12, 3) = 0.9 In excel we can use
=NORMINV(0.9, 12, 3)
which give 15.84465 weeks. Using the table look up, since the table gives
the probabilities up to 0, we need to first subtract 0.5 from 0.9 to get
0.4. Next we look up the table and find the closest number to 0.4 that
is 0.3997 which is attained at 1.28. Next, 1.28 is the standardized value
of the number of weeks: To get the number of weeks we need to “destandardize”: 1.28 = t−12
and thus t = 3 × 1.28 + 12 = 15.84 weeks.
3
3
October 4, 1999
Problem4: (15 points)
• The probability
Pr[−0.79 < Z < 1.07] = Pr[Z < 1.07] − Pr[Z < −0.79]
= (0.5 + value of table at 1.07) − [1 − Pr[Z < 0.79]
= 0.5 + value of table at 1.07 − (1 − 0.5 − value of table at 0.79
= 0.3577 + 0.2852
= 0.6429
Though the question asks you to use tables, I’ll mention the method using
Excel which is entering
=NORMSDIST(1.07)-NORMSDIST(-0.79)
and getting 0.642926.
• Using Excel you can enter
=NORMDIST(183,93,181,TRUE)-NORMDIST(-250,93,181,TRUE)
and get 0.661445. With tables you should first standardize both values:
z1 =
183 − 93
= 0.4972
181
and
z2 =
(−250) − 93
= −1.895028
181
Thus now we need the Pr[−1.895028 < Z < 0.4972] which is just like last
problem 0.4713+0.1915=0.6628.
Problem5: (15 points)
The mean of binomial distribution is µ = np = 75 ×√
0.7 = 52.5 and the variance
is σ2 = np(1 − p) = 52.5 × 0.25 = 13.125, and σ = 0.13125 = 3.62284. Thus,
since n ≥ 31 we can use normal approximation:
46 − 52.5
= −1.79417
3.62284
35 − 52.5
z2 =
= −4.83046
3.62284
z1 =
Thus we need to find Pr[−4.83046 < Z < −1.79417] for standard normal random
variable Z. This probability can be attained by Excel with
=NORMSDIST(-1.79417)-NORMSDIST(-4.83046)= 0.036392.
You could use table look up but since there is no entry for 4.83, it means
that the table value corresponding to it is effectively 0.5. So you may assume
the value of 0.5 for 4.83.
4
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