Question #280684

A random variable X ~ bino(5, 0.3). g(x) is a continuous and differentiable function on the interval (0, 5).


a. Write down the cumulative distribution function F(x) for the random variable X.

b. g(x) is an approximation of F(x). Construct g(x).



Expert's answer

a.

cumulative distribution function:

F(x)=P(Xk)=i=0k(ni)pi(1p)niF(x)=P(X\le k)=\displaystyle \sum_{i=0}^k\begin{pmatrix} n \\ i \end{pmatrix}p^i (1-p)^{n-i}

where k is successes in n independent Bernoulli trials

p is probability of success in trial

we have:

n=5,p=0.3n=5,p=0.3

then:

F(x)=P(Xk)=i=0k(5i)0.3i0.75iF(x)=P(X\le k)=\displaystyle \sum_{i=0}^k\begin{pmatrix} 5 \\ i \end{pmatrix}0.3^i 0.7^{5-i}


b.

g(0)=F(0)=P(X0)=0.75=0.1681g(0)=F(0)=P(X\le 0)=0.7^5=0.1681


g(1)=F(1)=P(X1)=P(0)+P(1)g(1)=F(1)=P(X\le 1)=P(0)+P(1)

P(1)=50.30.74=0.3602P(1)=5\cdot0.3\cdot0.7^4=0.3602

g(1)=0.1681+0.3602=0.5283g(1)=0.1681+0.3602=0.5283


g(2)=F(2)=P(X2)=P(0)+P(1)+P(2)g(2)=F(2)=P(X\le 2)=P(0)+P(1)+P(2)

P(2)=C520.320.53=0.1125P(2)=C^2_5\cdot0.3^2\cdot0.5^3=0.1125

g(2)=0.1681+0.3602+0.1125=0.6408g(2)=0.1681+0.3602+0.1125=0.6408


g(3)=F(3)=P(X3)=g(2)+P(3)g(3)=F(3)=P(X\le 3)=g(2)+P(3)

P(3)=C530.330.52=0.0675P(3)=C^3_5\cdot0.3^3\cdot0.5^2=0.0675

g(3)=0.6408+0.0675=0.6583g(3)=0.6408+0.0675=0.6583


g(4)=F(4)=P(X4)=g(3)+P(4)g(4)=F(4)=P(X\le 4)=g(3)+P(4)

P(4)=C540.340.5=0.0203P(4)=C^4_5\cdot0.3^4\cdot0.5=0.0203

g(4)=0.6583+0.0203=0.6786g(4)=0.6583+0.0203=0.6786


g(5)=F(5)=P(X5)=1g(5)=F(5)=P(X\le 5)=1


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