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Chapter 4: Q.4.17 (page 164)

Suppose that the distribution function of X given by

F(b)=0    b<0b4    0≤b<112+b−14    1≤b<21112    2≤b<31    3≤b

(a) Find P{X=i},i=1,2,3.

(b) Find P12<X<32.

Short Answer

Expert verified

(a) The value for the P{x=i} if i=1,2,3 are

P{1}=14

P{2}=16

P{3}=112

(b) The value for theP12<X<32is12.

Step by step solution

01

Given information (Part a)

The distribution function of X given as

F(b)=0    b<0b4    0≤b<112+b−14    1≤b<21112    2≤b<31    3≤b

02

Solution (Part a)

The calculation is given below,

P(X=1)=F(1+)−F(1−)

=12+1−14−14

=14

P(X=2)=F(2+)−F(2−)

=1112−12−2−14

=1112−12−14

=11−6−312

=16

Similarly,

P(X=3)=1−1112

=112

03

Final answer (Part a)

The value for the P{x=i} if i=1,2,3 are

P{1}=14

P{2}=16

P{3}=112

04

Given information (Part b)

Given in the question that

F(b)=0    b<0b4    0≤b<112+b−14    1≤b<21112    2≤b<31    3≤b

05

Solution (Part b)

The calculation is given below,

P12<X<32=F(3/2)−F(1/2)

=12+32−14−124

=12+12×14−12×14

=12

06

Final answer (Part b)

The value for theP12<X<32is12.

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Most popular questions from this chapter

Let N be a nonnegative integer-valued random variable. For nonnegative values aj, j Ú 1, show that

∑j=1∞a1+…+ajP{N=j}=∑i=1∞aiP[N≥i}

Then show that

E[N]=∑i=1∞P[N≥i}

and

E[N(N+1)]=2∑i=1∞iP{N≥i}

A box contains 5red and 5blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win \(1.10; if they are different colors, then you win -\)1.00. (That is, you lose $1.00.) Calculate

(a) the expected value of the amount you win;

(b) the variance of the amount you win.

Five distinct numbers are randomly distributed to players numbered1through 5. Whenever two players compare their numbers, the one with the higher one is declared the winner. Initially, players1and 2 compare their numbers; the winner then compares her number with that of player 3, and so on. Let X denote the number of times player 1 is a winner. FindPX=i,i=0,1,2,3,4.

A family has n children with probability αpn,n≥1whereα≤(1-p)/p

(a) What proportion of families has no children?

(b) If each child is equally likely to be a boy or a girl (independently of each other), what proportion of families consists of k boys (and any number of girls)?

Here is another way to obtain a set of recursive equations for determining Pn, the probability that there is a string of kconsecutive heads in a sequence of nflips of a fair coin that comes up heads with probability p:

(a) Argue that for k<n, there will be a string of kconsecutive heads if either

1. there is a string of kconsecutive heads within the first n-1flips, or

2. there is no string of kconsecutive heads within the first n-k-1flips, flip n-kis a tail, and flips n-k+1,…,nare all heads.

(b) Using the preceding, relate PntoPn-1. Starting with Pk=pk, the recursion can be used to obtain Pk+1, thenPk+2, and so on, up to Pn.
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