If a random variable X can have values of-3, -1, 2, and 5 with the respective pr

Question

If a random variable X can have values of-3, -1, 2, and 5 with the respective probabilities: 2k-3/10, K + 1/10, K - 1/10, and K - 2/10 Find distribution table of X, The mean(mu) and standard deviation (sigma) of X A fair coin is tossed until a head or five tails occurs. What is the expected number (E) of tosses of the coin to get those results? A fair coin is tossed three times. Let X = 0 if the first toss is a head and one if the first toss is a tail. Also let Y = total number of heads which occur, Find: Joint distribution table of X and Y, Covariance and correlation of X and Y, Are X and Y dependent or independent? Why.

Explanation / Answer

Given,

Outcomes(x)=-3,-1,2,5

Probabilities P(x)= 2k-3/10, k+1/10, k-1/10, k-2/10

we know, sum of all probabilities should always be 1

hence, (2k-3/10) + (k+1/10) + (k-1/10) + (k-2/10)=1

i.e, 5k-5=10

k=3

therefore the probabilities are given as,

3/10, 4/10, 2/10, 1/10

A.a) the distribution table for X is given as

x

P(x)

-3

0.30

-1

0.40

2

0.20

5

0.10

A.b) mean (µ) =x*P(x)

Standard Deviation(SD)=Sqrt [(x-µ)^2*P(x)]

both mean and SD can be calculated using the distribution table

x

P(x)

x*P(x)

(x-µ)^2

(x-µ)^2*P(x)

-3

0.30

-0.9

6.76

2.028

-1

0.40

-0.4

0.36

0.144

2

0.20

0.4

5.76

1.152

5

0.10

0.5

29.16

2.916

mean

-0.4

Variance

6.24

therefore, mean= -0.4

Standard Deviation(SD)= Sqrt(variance)

=sqrt(6.24)=2.498

Standard Deviation(SD)=2.498

A.2) we need to find expected number of tosses

possible outomces(X)=1,2,3,4,5 tosses

probabilities are

P(X=1)=50% or 0.5 [possibility of head occurring in first incident)

P(X=2)=tail occurring first and head occurring second

P(X=2)=0.5*0.5=0.25

similarly,

P(X=3)=0.5*0.5*0.5=0.125

P(X=4) =0.5*0.5*0.5*0.5=0.0625

P(X=5) =0.5*0.5*0.5*0.5*1=0.0625 (a head or tail in the fifth outcome will suffice our expectation)

now, the expected number is given as

E(x)=x*P(x)

E(x)=1*0.5+2*0.25+3*0.125+4*0.0625+5*0.0625

E(x)=1.9375

Hence the expected number is given as E(x)=1.9375

x

P(x)

-3

0.30

-1

0.40

2

0.20

5

0.10

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