Question #213452

1.verify that f(x)=2x/k(k+1) for x=0,1,2,.....k. can serve as a pmf of a random variable x

2.A fair coin is flipped until a head appears. Let N represent the number of tosses required to realise a head. Find the pmf of N

3.The pmf of a discrete random variable X is given by p(X=x)=kx for x=1,2,3,4,5,6.

Find the value of the constant k, p(X<4) and p(3<x<6)

4.A die is loaded such the probability that of a face showing up is proportional to the face number. Determine the probability of each sample point

5.Roll a fair die and let X be the square of the show up.Write the probability distribution of x hence compute p(X<15) and p(3<x<30)

6.Let x be a random variable the number of hours observed when two dice are rolled together once. show that x is a discrete random variable.

7.For each of the following determine c so that the function can serve as a pmf of a random variable X

  1. F(X)=cx for x=1,2,3,4,5
  2. f(x)=cx2 for x=0,1,2,...k
  3. f(x)=c(1/6)x for x=0,1,2,3,...
  4. f(x)=c2-x for x=0,1,2,....
1
Expert's answer
2021-07-16T16:28:46-0400

1.


f(x)=x(2k(k+1))0,x=0,1,2,...f(x)=x(\dfrac{2}{k(k+1)})\geq 0, x=0,1,2,...


x=0kf(x)=x=0kx(2k(k+1))\displaystyle\sum_{x=0}^kf(x)=\displaystyle\sum_{x=0}^kx(\dfrac{2}{k(k+1)})

=2k(k+1)(k(k+1)2)=1=\dfrac{2}{k(k+1)}(\dfrac{k(k+1)}{2})=1

f(x)=x(2k(k+1))0,x=0,1,2,...f(x)=x(\dfrac{2}{k(k+1)})\geq 0, x=0,1,2,... can serve as a pmf of a random variable x.x.


2.

A geometric random variable X with parameter p has probability mass function

f(N)=p(1p)N,N=0,1,2,....f(N) = p(1− p) ^N, N = 0,1,2,... .

Given p=0.5p=0.5


f(N)=0.5(10.5)N,N=0,1,2,....f(N) = 0.5(1− 0.5) ^N, N = 0,1,2,... .

f(N)=0.5N+1,N=0,1,2,....f(N) = 0.5 ^{N+1}, N = 0,1,2,... .

3.


k(1)+k(2)+k(3)+k(4)+k(5)+k(6)=1k(1)+k(2)+k(3)+k(4)+k(5)+k(6)=1

k=121k=\dfrac{1}{21}

P(X<4)=121(1+2+3)=27P(X<4)=\dfrac{1}{21}(1+2+3)=\dfrac{2}{7}

P(3<X<6)=121(4+5)=37P(3<X<6)=\dfrac{1}{21}(4+5)=\dfrac{3}{7}

4.

k+2k+3k+4k+5k+6k=1k+2k+3k+4k+5k+6k=1

k=121k=\dfrac{1}{21}

P(X=1)=121P(X=1)=\dfrac{1}{21}

P(X=2)=221P(X=2)=\dfrac{2}{21}

P(X=3)=17P(X=3)=\dfrac{1}{7}

P(X=4)=421P(X=4)=\dfrac{4}{21}

P(X=5)=521P(X=5)=\dfrac{5}{21}

P(X=6)=27P(X=6)=\dfrac{2}{7}

5.


x149162536p(x)1/61/61/61/61/61/6\begin{matrix} x & 1 & 4 & 9 & 16 & 25 & 36 \\ p(x) & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \end{matrix}

p(X<15)=16+16+16=12p(X<15)=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{1}{2}


p(3<X<30)=16+16+16+16=23p(3<X<30)=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{2}{3}

6.A discrete random variable is an rv whose possible values either constitute a

finite set or else can be listed in an infinite sequence in which there is a first

element, a second element, and so on (“countably” infinite).

A random variable the number of hours observed when two dice are rolled together once is

 countable number of possible values.

Therefore this random variable is discrete.


7.


1.

c+2c+3c+4c+5c=1c+2c+3c+4c+5c=1

c=115c=\dfrac{1}{15}

2.

x=0kcx2=ck(k+1)(2k+1)6=1\displaystyle\sum_{x=0}^{k}cx^2=\dfrac{ck(k+1)(2k+1)}{6}=1

c=1k(k+1)(2k+1)c=\dfrac{1}{k(k+1)(2k+1)}



3.

x=0c(16)x=c116=6c5=1\displaystyle\sum_{x=0}^{\infin}c(\dfrac{1}{6})^x=\dfrac{c}{1-\dfrac{1}{6}}=\dfrac{6c}{5}=1

c=56c=\dfrac{5}{6}

4.

x=0c(2)x=x=0c(12)x=c112=2c=1\displaystyle\sum_{x=0}^{\infin}c(2)^{-x}=\displaystyle\sum_{x=0}^{\infin}c(\dfrac{1}{2})^x=\dfrac{c}{1-\dfrac{1}{2}}=2c=1

c=12c=\dfrac{1}{2}



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