# elementary statistics i by housework

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```									                   DEPARTMENT OF MATHEMATICS AND STATISTICS
YORK UNIVERSITY
MATH 2560 3.0 , FALL 2005
ELEMENTARY STATISTICS I
ASSIGNMENT # 3
(SOLUTIONS)

QUESTION 1.
Ignoring twins and other multiple births, assume babies born at a hospital are independent events
with the probability that a baby is a boy and the probability that a baby is a girl both equal to 0.5.

a)        What is the probability that the next three babies are of the same sex?
SOLUTION: P = P(G and G and G)+P(B and B and B)= P(G)*P(G)*P(G) + P(B)*P(B)*P(B)=
= (0.5)*(0.5)*(0.5)+(0.5)*(0.5)*(0.5)= 0.25
b)        Define event B={at least one of the next two babies is a boy}. What is the probability
of the complement of event B?
SOLUTION: Event B{at least one of the next two babies is a boy}= {(B and G)or (G and B)or (B
and B)},
The complement of event B ={(G and G)}.
P(G and G)=0.5*0.5=0.25
c)        What is the probability that at least one of the next three babies is a boy?
SOLUTION: We will use the complement rule:
Event A is {at least one of the next three babies is a boy}.
P(A)=1-P(G and G and G)= 1-0.5*0.5*0.5= 0.875

QUESTION 2.
Part A.
Suppose there are three balls in a box. One of the balls is number 1, another is the number 2, and
the third is the number 3. You select two balls at random and without replacement from the box and
note the two numbers observed. Let X be the sum of the two balls selected.
a)       .       Draw the table of distribution for X.
SOLUTION: The sample space consists of six equally likely outcome {(1,2), (1,3), (2,1), (2,3),
(3,1), (3,2)}.
X={3, 4, 3, 5, 4, 5}.
Table of distribution of X:
X       3         4          5
P       1/3       1/3        1/3

b)             What is the probability that the sum is at least 4?
SOLUTION:             P( the sum is at least 4)= P(4)+P(5)= 2/3
c)             What is the mean of X?
SOLUTION:
Mean of X = 3*(1/3) + 4*(1/3) + 5*(1/3) = 4
Part B.
Let the random variable X be the number of repair calls that an appliance repair shop may
receive during an hour. The distribution of X is given below.
Value of X       0                1                 2                3                4
Probability      ???               0.3              0.12              0.18             0.2

a)        What is the value of the missing probability?
SOLUTION: P(X=0)=1- (0.3 + 0.12 + 0.18 + 0.2) = 0.2
b)        What is the probability that the repair shop receives at least three repair calls during
an hour?
SOLUTION: P(at least three calls)= P(X=3) + P(X=4) = 0.18 + 0.2 = 0.38
c)        What is the expected number of repair calls during an hour?
SOLUTION: The expected number of repair calls = mean of X=
= 0*0.2 + 1*0.3 + 2*0.12 + 3*0.18 + 4*0.2 = 1.88 call per hour

QUESTION 3.
When figure skaters need to find a partner for “pair figure skating”, it is important to find a
partner who is compatible in weight. The weight of figure skaters can be modeled by a
normal distribution. For male skaters, the mean is 170 lbs. with a standard deviation of 10
lbs. For female skaters, the mean is 110 lbs. with a standard deviation of 5 lbs. Let the
random variable X=the weight of female skaters and the random variable Y=the weight of
male skaters.
a)    The weight of a pair of figure skaters (a male and a female) can be thought of as a new
random variable. Let the random variable W=X+Y. What is the mean of this new random
variable W?
SOLUTION: Mean (W)=Mean(X)+Mean(Y)=170+110=280 lbs.
b)    Suppose we consider the weight of the male partner and the female partner to be
independent. What is the standard deviation of the random variable W?
SOLUTION: Var(X)=100; Var(Y)=25; Var(W)=Var(X)+Var(Y)=125;
 w  125  11.18 lbs.
c) It does not seem likely that the weight of the male partner and the female partner would be
independent. If the correlation between X and Y equals 0.77, what is the standard deviation
of the random variable W?
SOLUTION:
Var(W )  Var( X )  Var(Y )  2  x y  100  25  2 * 0.77 *10 * 5  202;
 w  202  14.21lbs

QUESTION 4.
Consider two events: E and F. We know that P(E)=P(F)=0.7.
a)  Are the two events E and F disjoint?
SOLUTION: No; P(E)+P(F)=1.4;
b)  If E and F were known to be independent, what is the probability that they occur
simultaneously?
SOLUTION: P(E and F)=P(E|F)*P(F)=P(E)*P(F)=0.7*0.7=0.49
c)  If E and F were known to be independent, what is the probability that at least one of them
occurs?
SOLUTION: P(E or F)=P(E)+P(F)-P(E and F)=0.7+0.7-0.49=0.91;
d)  Suppose we know that P(F|E)=0.9. What is the probability that at least one of them occurs?
SOLUTION: P(E or F)=P(E) + P(F) - P(E and F) = P(E) + P(F) - P(E|F)*P(F) = 1.4 - 0.9*0.7 =
0.77
e)Suppose we know that P(F|E)=0.9. What is P(E|F)?
SOLUTION: P(E|F)=P(E and F)/P(F)=P(F|E)*P(E)/P(F)=(0.9*0.7)/0.7 =0.9

QUESTION 5
The table below shows the political affiliation of 1,000 random selected American voters
and their positions on the school of choice program.
Position                  Democrat                   Republican               Other
Favor                     260                        120                      240
Oppose                    40                         240                      100

Let the event D={voter is a Democrat}, R={voter is a Republican}, and F={voter favors the school
of choice program}.For each of the following questions, write the probability in symbols
(e.g., P(D)) and calculate the probability.

a)  What is the probability that a randomly selected voter favors the school of choice program?
SOLUTION: 620/1000=0.62
b)  What is the probability that a randomly selected Republican favors the school of choice
program?
SOLUTION: 120/360=1/3=0.33…
c)  What is the probability that a randomly selected voter who favors the school of choice
program is a Democrat?
SOLUTION: 260/620=0.42

QUESTION 6.
A prisoner has just escaped from jail. There are three roads leading away from the jail. If the
prisoner selects road A to make good her escape, the probability that she succeeds is 0.25. If
she selects road B, the probability that she succeeds is 0.2. If she selects road C, the
probability that she succeeds is (1/6). Furthermore, the probability that she selects each of
these roads is the same. It is (1/3). If the prisoner succeeds in her escape, what is the