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.

1
Expert's answer
2022-03-18T01:14:35-0400

n=20, p=0.1, q=1p=10.1=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)pkqnk,P(X=k)=\begin{pmatrix} n \\ k \end{pmatrix}\cdot p^k \cdot q^{n-k},where (nk)=n!k!(nk)!\begin{pmatrix} n \\ k \end{pmatrix}=\cfrac{n! } {k! \cdot(n-k)! }

is the binomial coefficient.

P(X=k)==(20k)0.1k0.9nk==20!k!(20k)!0.1k0.9nk.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(X4)=P(X=0)+P(X=1)++P(X=2)+P(X=3)++P(X=4)==20!0!20!0.100.920++20!1!19!0.110.919++20!2!18!0.120.918++20!3!17!0.130.917++20!4!16!0.140.916==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=200.1=2.\mu=np=20\cdot0.1=2.

The standard deviation


σ=npq=200.10.9=1.34.\sigma=\sqrt{npq} =\sqrt{20\cdot0.1\cdot0.9} =1.34.














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