A coin is tossed repeatedly; x = number of the toss at which a head first appears.

Short Answer

Expert verified

The required values are mentioned below.

μ=2var(x)=2σ=2

Step by step solution

01

Given Information

A coin is tossed repeatedly.

02

Definition of the cumulative distribution function

The likelihood that a comparable continuous random variable has a value less than or equal to the function's argument is the value of the function.

03

Find the values.

The random variables are given below

x=1px1=px=2px2=pq

Solve further.

x=3px3=pq2x=4px4=pq3

Solve further.

x=5px5=pq4x=npxn=pqn1

The mean is given below.

μ=x=1xpqx1=p1+2q+3q2+4q3+5q4+

LetS=1+2q+3q2+4q3+5q4++

qS=q+2q2+3q3+4q4+5q5+SqS=1+q+q2+q3+q4++S(1q)=11qS=1(1q)2

The mean becomes as follows.

μ=p(1q)2μ=1/2(1/2)2μ=2

The variance is given below.

var(x)=(x2)2pqx1=(x1)qx=q+q3+4q4+9q5+=q+q31+4q+9q2+16q3+

LetS1=1+4q+9q2+16q3+

S1=1+q(1q)3

The value of variance becomes as follows.

var(x)=q+q31+q(1q)3var(x)=12+×3/2(0.5)3var(x)=2

The standard deviation is given below.

σ=var(x)σ=2

Cumulative function is given below.

x=1Fx1=1/2x=2Fx2=3/4

Solve further.

x=3Fx3=7/8x=4Fx4=15/16

Solve further.

x=5Fx5=31/32

The graph is shown below

Hence, the required values are mentioned be

μ=2var(x)=2σ=2

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