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1. Course: Business Mathematics (1429)
Semester: Autumn, 2022
ASSIGNMENT No. 1
Q. 1 (a)A major urban hospital has gathered data on the number of heart
attack victims seen. The given table indicates the probabilities of
different numbers of heart attack victims being treated in the emergency
room on a typical day.
Number of
Victims
P(n)
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Treated (n)
Fewer Than 5 0.08
5 0.16
6 0.30
7 0.26
More than 7 0.20
i. What is the probability that 5 or fewer victims will be seen?
Probability = 0.16 + 0.08
Probability = 0.24
ii. What is the probability that 5 or more victims will be seen?
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Probability = 0.16 + 0.30 + 0.26 + 0.20
Probability = 0.92
iii. What is the probability that no more than seven victims will be
seen?
Probability = 0.26
(b) The sample space of an experiment consists of five simple events: E1,
E2, E3, E4 and E5. These events are mutually exclusive. The probabilities of
occurrence of these events are P(E1) = .20, P(E2) = .15, P(E3). 25, P(E4) = .30,
P(E5) =.10. Several compound events can be defined for this experiment. They
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9. Q. 2 (a)
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a. A random variables X has the following .
X -2 -1 0 1 2 3
P(x) 0.1 k 0.2 2k 0.3 3k
Find
(i) k
Sum of p(X) = 1
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p(-2< X <2) = 3(0.067)
p(-2< X <2) = 0.201
(v) p(X < 1)
p(X < 1) = p(X=0)+p(X=-1)+p(X=-2)
p(X < 1) = 0.2 + k + 0.1
p(X < 1) = 0.2 + 0.067 + 0.1
p(X < 1) = 0.367
(b) Construct the discrete probability distribution which corresponds to the
experiment of the single roll of a pair of dice. Assume equal likelihood of
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occurrence of each side of a die and let the random variable X equal the sum
of the dots which appear on the pair
since a dice has 6 sides , here we are taking a pair so the total possibilities are 36,
6×6=36 ..
let x be the sum of dots on the up side of dice then the total 36 possibilities are ;
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) . . .. (2,6)
(3,1) . . .. (3,6)
(4,1) . . . . . (4,6)
(5,1). . . . .. (5,6)
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(6,1) . . . . .. (6,6)
when x(sum) = 2 there is 1 dot on each dice. the possibility is 1 as (1,1) come only
once in 36 possibilities, thus the probability is 1/36 ..
similarly when x=3 the possibilities are 2
as (1,2) and (2,1) that adds upto give sum 3 .. the probability is 2/36=1/18
continue the same process upto x=12 . and you'll get all the probabilities.
Q. 3
(a) Sketch the plane representing 3x + 0y + 8z = 5
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15. (b) Give the points (–4, 8) and (6, –12):
(i)Determine the midpoint of the line segment connecting the points.
(ii) Determine the distance separating the two points.
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17. Q. 4
(a) The book value of a machine is expressed by the function: V=25,000 –
1,225t Where V equal the book value in dollars and t equals the age of the
machine expressed in years.
1. Identify the t and V intercepts.
V = 0 then t = 20.40
t = 0 then V = 25000
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2. Interpret the meaning of the intercepts.
The V−intercept and t−intercepts are 25000 and 20.40 respectively.
3. Interpret the meaning of the slope.
V=25,000 – 1,225t
V = -1225t + 25000
V = mt + c
m = -1225
-1225 is the slope.
(b) Solve the following inequalities:
i. |x2– 2| ≥ 2
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19. ii. |6t – 15| ≤ – 6
Q. 5
(a) Average Starting salaries for students majoring in business have been
increasing. The equation which predicts average starting salary is s = 20,250
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+ 1,050t where s equals average starting salary and t is time measured in
years since 1990 (t=0)
i. Identify the s and t intercepts for this equation.
equation is S = 20,250 + 1,050 * t
s is the average starting salary.
t is the number of years from 1990.
t = 0 references the year 1990.
t = 1, 1991, etc.
the S intercept is the value of S when t = 0.
that makes the S intercept 20,250.
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ii. Intercepts these values where they are meaningful.
The equation tells you what the average starting salary is for students majoring in
business, by the number of years from 1990. in 1990, the average starting salary is
20,250.
In 2000, the average starting salary is
20,250 + 1,050 * 10 * 1,050 = 30,750.
In 2020, the average starting salary is 20,250 + 1,050 * 30 = 51,750.
Here's a graph that shows the average starting salary every 5 years from 1990 to
2000.in the graph equation, y takes the place of S and x takes the place of t.
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22. (b) Construct the discrete probability distribution which corresponds to the
experiment of tossing a fair coin three times. Let the random variables X equal
the number of heads occurring in three tosses, what is the probability of two or
more heads?
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First, let's think about what the probability of tossing a head or tail on one coin is.
That would be 0.5 since the coin is "fair."
How about tossing three heads successively (HHH)? That would be 0.5 x 0.5 x
0.5, or (0.5)3
How about tossing two heads and then one tail in that order (HHT)? That would
be 0.5 x 0.5 x 0.5, or (0.5)3
How about tossing two tails and then one head in that order (TTH)? That would
again be (0.5)3.
Doing more of these examples, you can convince yourself that the probability of
getting a particular outcome tossing three coins in a row is (0.5)3.
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Let us now write out all the possible outcomes from tossing 3 coins in a row:
{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
In other words, there are 8 possible outcomes, and each outcome has a probability
of (0.5)3
If X represents the number of heads that will result from the 3 tosses, X can be 0,
1, 2, or 3 based on the above outcome (that is, 0 heads can result, 1 head can
result, 2 head can result or 3 head can result).
Now let's consider these questions:
What is the probability that X = 0? That would correspond to the outcome TTT,
which would be (0.5)3
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What is the probability that X = 1? That would correspond to the outcomes HTT,
THT, or TTH, depending on when the head comes. That would be (0.5)3 + (0.5)3 +
(0.5)3 or 3(0.5)3
A probability distribution is basically all the possible values of the random
variable and their corresponding probabilities. So for our case, this would be the
probability distribution:
X = 0, (0.5)3
X = 1, 3(0.5)3
X = 2, 3(0.5)3
X = 3, (0.5)3
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The sum of all probabilities in a probability distribution should be 1, which is true
in our case.
Graphing this would be simple. Since this is a discrete function, you should have
to denote each value of x and its probability using vertical lines.
E(X) is basically the expectant value, or weighted average, of the probability
distribution. To calculate the E(X), we need to sum x*p(x) for all values of x. So
in this case, it would be 0*(0.5)3 + 1*3(0.5)3 + 2*3(0.5)3 + 3*(0.5)3, which is 1.5
V(X) is the variance and is a measure of how far each random variable is from the
mean.
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To calculate the variance we need to average the square of the difference of each
random variable and the mean. So, in other words, V(X) = E[(X-E(X))2]
So, (0-1.5)2 (0.5)3 + (1-1.5)2 3(0.5)3 + (2-1.5)2 3(0.5)3 + (3 -1.5)2 (0.5)3 = 0.75
Then we divide 5 by the number of trials, which in this case was 3 (since we
tossed the coin 3 times). So 5/3 is the variance
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