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Let Ten be the damage incurred (in $) in a sure type of accident during a given year. Possible X values are 0, g, 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 impairment due to certain blazon of accident in a provided year.

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

For X=0, and so

Y=10+100

FOR X=1000, 5000 and 10000, and so

Y=Ten-500+100

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

The below tabular array represents the probability distribution of Y as,

X 0 1000 5000 10000
Y 100 600 4600 9600
P(y) 0.8 0.one 0.08 0.02

The premium charged is obtained as

Due east(Y)=y * P(y)

=100*0.eight+600*0.1+4600*0.08+9600*0.02

=$700

The chargeable premium amount is $700

Possible values of 10, the number of components in a system submitted for repair that must exist replaced, are ane,2, 3, and iv with corresponding probabilities .15, .35, .35, and .fifteen, respectively. a. Calculate E(Ten) 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 – Ten)]? [Notation: It is non generally truthful that Eastward(c/Y) = c/E(Y)]

(a)

Use the formula for mean as follows:

µ = Eastward (X)

=∑four 10=1 x * P(x)

=1*0.15 +2 * 0.35 + …+4* 0.15

=2.5

The mean number of the component in a organization submitted for repair is found to be 2.5. The expected value E(5-10) can be calculated by using the law of linearity of expectation. The linearity belongings of expectation suggests:

E(5-X) = v – (X)

=5 – 2.5

=2.5

Part a

The expected values are found to exist:

East(X) = ii.5

E(5 – X) = ii.5

Explanation | Hint for next step

The hateful number of components submitted for repair is establish to be 2.five and the expected value for E(5 – X) is found to exist ii.five

Explanation | Hind for the side by side step

The hateful number of components submitted for repair is found to be ii.5 and the expected value for Eastward(5 – Ten) is found to exist 2.5.

(b)

The modified distribution is as shown in the table:

Y P(Y)
37.five 0.15
l 0.35
75 0.35
150 0.xv

The expected value is:

E(Y) = 37.5*0.15 + l*0.35 + …+ 150*0.xv

=71.875

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

Office b

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

Explanation

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

Part a

The expected values are found to be:

Eastward(X) = 2.5

Due east(v – X) = 2.5

Part b

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

A chemical supply visitor currently has in stock 100 lb of a certain chemical, which information technology 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 iii iv
p(x) .2 .four .3 .1

Compute E(Ten) and V(X). And so compute the expected number of the pounds left after the side by side 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 Ten.]

The random variable X is divers as

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

The visitor has in stock 100 lb of the chemic

The company sells the chemical in lots v lb

The pmf of X is given by:

10 1 2 iii iv
P(x) 0.two 0.iv 0.3 0.ane

The mean value X is

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

= 1(0.2) + 2(0.4) + three(0.3) + four(0.1)

=0.two + 0.eight +0.ix+0.4

=ii.3

The variance of X, denoted by 5(Ten) or just σtwo , is

Five(Ten) = ∑(X-µ)two.p(x)

=∑(x-2.3)2 . p(x)

=(ane-2.3)2 .(0.2) + (2-2.iii)2 .(0.4)+(iii-2.3)2 .(0.3) + (four -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 announce the number of pounds of the chemical left with the company afterwards theext cstomer's order is shipped

The number of pounds left later the side by side customer'due south society is shipped is equal to the total stock with the company minus 5 times the number of lots ordered past the customers. (we multiply the number of lots by 5 since the lots are sold to customer in five-lb lots).

In other words, Y = 100 – 5X

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

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

So the expectation of Y can be computed as follows:

East(Y) = E(100 – 5X)

= 100 – 5E(10)

=100 – 5*2.three

=88.five

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

Five(aX + b) = a2Five(X)

Then the variance of Y can be computed equally follows:

5(Y) = 5(100 – 5X)

=v2 * V(X)

=25*0.81

20.25

Q48

NBC News reported on May two, 2022, that i in 20 children in the The states has a nutrient allergy of some sort. Consider selecting a random sample of 25 children and let X be the number in the sample who have a nutrient allergy. And then X⁓Bin(25,.05).

  1. Determine both P(X≤three) and P(X<3).
  2. Determine P(X≥ 4).
  3. Determine P(ane ≤ X ≤ 3).
  4. What are E(X) and σ10 ?
  5. In a sample of 50 children, what is the probability that none has a food allergy?

Let Ten 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 (Ten = x) = 25Cten(0.05)10 (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 = one) + P (Ten = 2) + P (X =3)

=[ 25c0(0.005)0 (one – 0.05)25-0+25cone(0.05)1(i-0.05)25-1+25c2(0.05)two (ane-0.05)25-two + 25c3(0.05)three(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 three is,

P( X<3) = P (10=0) + P(Ten=one) + P(X=2)

=[25c0(0.05)0(1-0.05)25-2+ 25c1(0.05)1(1-0.05)25-1+ 25c2(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( Ten≥4) = ane- P(10 < 4)

= 1 –P( 10≤ 3)

=1 – 0.9659

©

The value of P( 1 ≤ X ≤ 3) is,

= [25cone(0.05)one(1-0.05)25-i+ 25c2(0.05)2(ane-0.05)25-2+ 25c3(0.05)three(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)

=i.25 the standard deviation of the random variable X is,

= 1.09

(due east)

The probability that none of the fifty Children'southward are food allergy is,

P(Ten = 0) = 50c0(0.05)0(ane- 0.05)50 – 0

= ane * 1 * (0.95)l

= 0.0769

Therefore, the probability that none of the l children's are nutrient allergy is 0.0769

50. A particular telephone number is used to receive both vocalism calls and fax letters. 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 half-dozen of the calls involve a fax message?

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

c. At to the lowest degree vi of the phone call involve a fax bulletin?

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

  • On the basis of the provided information, the full number of incoming calls (northward=25) and the total probability of receiving incoming calls involving fax letters (p=025)

The calculation of the probability that at most half-dozen of the calls involve a fax bulletin is:

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

= 0.5611

Part a

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

Explanation | Common mistakes | Hint for side by side step

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

(b)

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

= 0.1828

Part b

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

Explanation | Common error| Hint for adjacent step

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

(c)

The adding of the probability that more than 6 of the calls involve a fax bulletin is:

P (k > 6) = 1-P(k ≤ six) + P (chiliad = 6)

= one – 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

Virtually about 62.17% of the calls involve a fax bulletin 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 bulletin is :

P (k > half dozen) = 1 – P(thousand ≤ half-dozen)

= one- 0.5611

= 0.4389

Part d

The probability that more than than half dozen of the calls involve a fax bulletin is 0.4389

Explanation | common mistake | Hint for adjacent step

Almost 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 ii female person children. What is the pmf of Ten = the total number of male children born to the brothers? What is East(Ten), and how does it compare to the expected number of male children built-in to each brother?

The known experiment follows a negative binomial distribution.

  1. Each family needs ii female children, therefore, the total number of female children, X, in 3 families is r = 2+ 2+ 2 =half-dozen
  2. The probability of success remains abiding for each contained trail,

P(female) = P(male)

=one/ii

=0.5

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

Notice the probability mass office of X

Using r =  half dozen and p = 0.5, the probability mass role of X is,

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

=nb(ten ; 6, 0.5)

The expected value of the negative binomial distribution is,

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