Question #313098

Let x be a binomial random variable with n=20 and p = 0.1.


a. Find the formula for the probability distribution of x.


b. Calculate P(X≤4)


c. Calculate the mean and standard deviation of X.

Expert's answer

n=20, p=0.1, q=1−p=1−0.1=0.9.n=20, \ p=0.1,\ q=1-p=1-0.1=0.9.


a. The probability of getting exactly k successes in n independent Bernoulli trials is given by the probability mass function:

P(X=k)=(nk)⋅pk⋅qn−k,P(X=k)=\begin{pmatrix} n \\ k \end{pmatrix}\cdot p^k \cdot q^{n-k},where (nk)=n!k!⋅(n−k)!\begin{pmatrix} n \\ k \end{pmatrix}=\cfrac{n! } {k! \cdot(n-k)! }

is the binomial coefficient.

P(X=k)==(20k)⋅0.1k⋅0.9n−k==20!k!⋅(20−k)!⋅0.1k⋅0.9n−k.P(X=k)=\\=\begin{pmatrix} 20 \\ k \end{pmatrix}\cdot 0.1^k \cdot 0.9^{n-k} =\\ =\cfrac{20! } {k! \cdot(20-k)! } \cdot 0.1^k \cdot 0.9^{n-k} .


b.

P(X≤4)=P(X=0)+P(X=1)++P(X=2)+P(X=3)++P(X=4)==20!0!⋅20!⋅0.10⋅0.920++20!1!⋅19!⋅0.11⋅0.919++20!2!⋅18!⋅0.12⋅0.918++20!3!⋅17!⋅0.13⋅0.917++20!4!⋅16!⋅0.14⋅0.916==0.9568.P(X\le4) =\\P(X=0)+P(X=1)+\\+P(X=2)+P(X=3)+\\+P(X=4)=\\ =\cfrac{20!}{0!\cdot20!}\cdot0.1^0\cdot0.9^{20} +\\+ \cfrac{20!}{1!\cdot19!}\cdot0.1^1\cdot0.9^{19} +\\+ \cfrac{20!}{2!\cdot18!}\cdot0.1^2\cdot0.9^{18} +\\+ \cfrac{20!}{3!\cdot17!}\cdot0.1^3\cdot0.9^{17} +\\+ \cfrac{20!}{4!\cdot16!}\cdot0.1^4\cdot0.9^{16} =\\ =0.9568.


c. The mean


μ=np=20⋅0.1=2.\mu=np=20\cdot0.1=2.

The standard deviation


σ=npq=20⋅0.1⋅0.9=1.34.\sigma=\sqrt{npq} =\sqrt{20\cdot0.1\cdot0.9} =1.34.














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