#### 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

V(X) = ∑(X-µ)^{2}.p(x)

=∑(x-2.3)^{2} .
p(x)

=(1-2.3)^{2}
.(0.2) + (2-2.3)^{2} .(0.4)+(3-2.3)^{2} .(0.3) + (4 -2.3)^{2}
.(0.1)

=1.69(0.2) + 0.18(0.4)+1.47(0.3)+11.56(0.1)

=0.81

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)

=[ ^{25}c_{0}(0.005)^{0}
(1 – 0.05)^{25-0}+^{25}c_{1}(0.05)^{1}(1-0.05)^{25-1}+^{25}c_{2}(0.05)^{2}
(1-0.05)^{25-2} + ^{25}c_{3}(0.05)^{3}(1-0.05)^{25-3}]

=0.2774 + 0.3650+ 0.2305 + 0.0930

= 0.9659

The probability of having food allergy in the samples of less than 3 is,

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

=[^{25}c_{0}(0.05)^{0}(1-0.05)^{25-2}+
^{25}c_{1}(0.05)^{1}(1-0.05)^{25-1}+ ^{25}c_{2}(0.05)^{2}(1-0.05)^{25-2}]

=0.8729

(b)

The probability of having food allergy in the samples of greater than or equal to 4 is,

P( X≥4) = 1- P(X < 4)

= 1 –P( X≤ 3)

=1 – 0.9659

©

The value of P( 1 ≤ X ≤ 3) is,

^{= }[^{25}c_{1}(0.05)^{1}(1-0.05)^{25-1}+
^{25}c_{2}(0.05)^{2}(1-0.05)^{25-2}+ ^{25}c_{3}(0.05)^{3}(1-0.05)^{25-3}]

= 0.3650 + 0.2305 + 0.0930

= 0.6885

(d)

The mean of the randon variable X is,

E (X) = np

= 25(0.05)

=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:

= 0.0008 + 0.0063 + 0.0251 + 0.0641 + 0.1175 + 0.1645 + 0.1828

= 0.5611

Part a

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.