A bottle makes and sells plastic bottles. It is observed that when the price is $42, only 12 bottles are sold in a week; but when the price decreases to $10, weekly sales rise to 25.

Assuming that demand can be modelled by a linear function,

(a) Obtain a formula for P in terms of Q;

(b) Sketch a graph of P against Q;

(c) Comment on the likely reliability of the model

Exercise 3: (14 Points)

A company sells packs of cell phone covers for $2. If a customer orders more than 100 packs, the company is prepared to reduce the unit price by 10 cents for each pack bought above 100 up to a maximum of 350 in a single order.

(a) How much does it cost to buy 140 packs?

(b) If the cost is $3524, how many packs are ordered?

Let X be the damage incurred (in
$) in a certain type of accident during a given year. Possible X values are 0,
1000, 5000, and 10000 with probabilities .8 .1 .08 and .02 respectively. a
particular company offers a $500 deductible policy. if the company wishes its
expected profit to be $100 what premium amount should it charge?

Let X be the variable that represents the damage due to certain type of accident in a provided year.

As per information provided,
the deductible amount is $500 and the expected premium charge is $100, the
premium function can be defined as:

For X=0, then

Y=X+100

FOR X=1000, 5000 and 10000,
then

Y=X-500+100

The chargeable amount is
obtained by substituting the values of X in the premium function.

The below table represents the probability distribution of Y as,

X

0

1000

5000

10000

Y

100

600

4600

9600

P(y)

0.8

0.1

0.08

0.02

The premium charged is
obtained as

E(Y)=y * P(y)

=100*0.8+600*0.1+4600*0.08+9600*0.02

=$700

The chargeable premium
amount is $700

Possible values of X, the number of components
in a system submitted for repair that must be replaced, are 1,2, 3, and 4 with
corresponding probabilities .15, .35, .35, and .15, respectively. a. Calculate
E(X) and then E(5 – X). b. Would the repair facility be better off charging a
flat fee of $75 or else the amount $[150/(5 – X)]? [Note: It is not generally
true that E(c/Y) = c/E(Y)]

(a)

Use the
formula for mean as follows:

µ = E (X)

=∑^{4}_{x=1
}x * P(x)

=1*0.15 +2
* 0.35 + …+4* 0.15

=2.5

The mean number of the component in a system submitted for repair is found to be 2.5. The expected value E(5-X) can be calculated by using the law of linearity of expectation. The linearity property of expectation suggests:

E(5-X) = 5
– (X)

=5 – 2.5

=2.5

Part a

The expected values are
found to be:

E(X) = 2.5

E(5 – X) = 2.5

Explanation | Hint for next step

The mean number of components submitted
for repair is found to be 2.5 and the expected value for E(5 – X) is found to
be 2.5

Explanation | Hind for the next step

The mean number of components submitted for repair is found to be 2.5 and the expected value for E(5 – X) is found to be 2.5.

(b)

The modified distribution
is as shown in the table:

Y

P(Y)

37.5

0.15

50

0.35

75

0.35

150

0.15

The expected value is:

E(Y) = 37.5*0.15 + 50*0.35 + …+ 150*0.15

=71.875

The expected repair charge by the repair
person is found to be 71.875. Decide which condition is better for repair
person: the variable cost or the fixed cost. The variable cost implies that he
would make on average $71.88. The fixed cost method says that he should charge
$75. It is profitable for him to charge a flat fee of $75 for repair.

Part b

The repair facility should charge a flat fee of
$75.

Explanation

The variable cost of charging strategy indicates that on average the repair person would make $71.88 per repair. The repair facility should charge a flat fee of $ 75.

Part a

The expected values are found to be:

E(X) = 2.5

E(5 – X) = 2.5

Part b

The repair facility should charge a flat
fee of $75.

A chemical supply company
currently has in stock 100 lb of a certain chemical, which it sells to
customers in 5-lb latches. Let X = the number of batches ordered by a randomly
chosen customer, and suppose that X has pmf

X

1

2

3

4

p(x)

.2

.4

.3

.1

Compute E(X) and V(X). Then compute the
expected number of the pounds left after the next customer’s order is shipped
and the variance of the number of pounds left. [Hint: The number of pounds left
is a linear function of X.]

The random variable X is defined as

X = the number of lots ordered by a randomly
chosen customer

The company has in stock 100 lb of the chemical

The company sells the chemical in lots 5 lb

The pmf of X is given by:

X

1

2

3

4

P(x)

0.2

0.4

0.3

0.1

The mean value X is

E(X) = µ = 1.P(1) + 2.p(2) + 3.p(3) +4.p(4)

= 1(0.2) + 2(0.4) + 3(0.3) + 4(0.1)

=0.2 + 0.8 +0.9+0.4

=2.3

The variance of X,
denoted by V(X) or just σ^{2} , is

Let Y denote the number
of pounds of the chemical left with the company after theext cstomer’s order is
shipped

The number of pounds left
after the next customer’s order is shipped is equal to the total stock with the
company minus 5 times the number of lots ordered by the customers. (we multiply
the number of lots by 5 since the lots are sold to customer in 5-lb lots).

In other words, Y = 100 –
5X

We know that the
expectation of a linear function aX+b is given by:

E(aX + b) = a.E(X) + b

So the expectation of Y
can be computed as follows:

E(Y) = E(100 – 5X)

= 100 – 5E(X)

=100 – 5*2.3

=88.5

We know that the variance of a linear function aX + b is given by:

V(aX + b) = a^{2}V(X)

So the variance of Y can
be computed as follows:

V(Y) = V(100 – 5X)

=5^{2} * V(X)

=25*0.81

20.25

Q48

NBC News reported on May 2, 2013, that 1 in 20 children in the United States has a food allergy of some sort. Consider selecting a random sample of 25 children and let X be the number in the sample who have a food allergy. Then X⁓Bin(25,.05).

Determine both
P(X≤3) and P(X<3).

Determine P(X≥
4).

Determine P(1
≤ X ≤ 3).

What are E(X)
and σ_{x} ?

In a sample of
50 children, what is the probability that none has a food allergy?

Let X be the number in the sample who has a food allergy and it follows the binomial distribution with the parameters (n=25, p = 0.05).

The probability mass function of
X is,

P (X = x) = ^{25}C_{x}(0.05)^{x}
(1-0.05)^{25-x }; x = 0,1,2,3,…,25

(a)

The probability of having food allergy in the
number of samples less than or equal to 3 is,

P(X ≤ 3) = P (X = 0) + P(X = 1) + P (X = 2) + P
(X =3)

=1.25 the standard deviation of the random
variable X is,

= 1.09

(e)

The probability that
none of the 50 Children’s are food allergy is,

P(X = 0) = 50_{c0}(0.05)^{0}(1-
0.05)^{50 }^{– 0}

^{= 1 * 1 * (0.95)}50

= 0.0769

Therefore, the probability that none of the 50
children’s are food allergy is 0.0769

50. A particular
telephone number is used to receive both voice calls and fax messages. Suppose
that 25% of te incoming calls involve
fax mesasges, and consider a sample of 25 incoming calss. What is the
probability that

a. At most 6 of the calls
involve a fax message?

b. Exactly 6 of the calls
involve a fax message?

c. At least 6 of the call
involve a fax message?

d. Moe than 6 of the
calls involve a fax message?

On the basis
of the provided information, the total number of incoming calls (n=25) and the
total probability of receiving incoming calls involving fax messages (p=025)

The calculation of the probability that at most 6 of the calls involve a fax message is:

The probability that at most 6 of the calls
involve a fax message is 0.5611.

Explanation | Common mistakes | Hint for next
step

Nearly about 56.11% of calls involve fax message
when the total number of incoming calls is 25.

(b)

The calculation of the probability that exactly 6 of the calls involve a fax message is:

= 0.1828

Part b

The probability that exactly 6 of the calls
involve a fax message is 0.1828.

Explanation | Common mistake| Hint for next step

Nearly about 18.28% of the calls involve a fax message when the total number of incoming calls is 25.

(c)

The calculation of the probability that more than 6 of the calls involve a fax message is:

P (k > 6) = 1-P(k ≤ 6) + P (k = 6)

= 1 – 0.5611 + 0.1828

= 0.6217

Part c

The probability that at least 6 of the calls involve a fax message is 0.6217.

Explanation | Hint for the next step

Nearly about 62.17% of the calls involve a fax message when the total number of incoming calls is 25.

(d)

The calculation of the probability that more than
6 of the calls involve a fax message is :

P (k > 6) = 1 – P(k ≤ 6)

= 1- 0.5611

= 0.4389

Part d

The probability that more than 6 of the calls
involve a fax message is 0.4389

Explanation | common mistake | Hint for next step

Nearly about 43.89% of the calls involve a fax
message when the total number of incoming calls is 25.

77. Three brothers and their wives decide to have children until each family has two female children. What is the pmf of X = the total number of male children born to the brothers? What is E(X), and how does it compare to the expected number of male children born to each brother?

The known experiment follows a negative binomial distribution.

Each family needs two female children, therefore, the total number of female children, X, in three families is r = 2+ 2+ 2 =6

The probability of success remains constant for each independent trail,

P(female) = P(male)

=1/2

=0.5

The experiment
is performed until each family has two female children.

Find the probability mass function of X

Using r =
6 and p = 0.5, the probability mass function of X is,

P(X) = nb( X; r,p)

=nb(x ; 6, 0.5)

The expected value of the negative binomial distribution is,

Therefore, an expected
number of male children born to each brother is 6.