Question #105111

Let a random variable follow the binomial distribution with parameters n =13, p =0.1. Compute the probability of x ≤4.


1
Expert's answer
2020-03-11T11:06:49-0400

μ is the binomial random variable,μB(n,p).μ is the number of successes in n Bernoulli trials.P{μ=k}=Cnkpk(1p)nk,k=0,1,,n.Then P{μk}=i=0kCnipi(1p)ni.μB(13,0.1).P{μ4}=i=04C13i(0.1)i(0.9)13i=(0.9)13+13(0.1)(0.9)12+78(0.1)2(0.9)11+2213(0.1)3(0.9)10+51113(0.1)4(0.9)90.9935.\mu\text{ is the binomial random variable}, \mu\in B(n,p).\\ \mu\text{ is the number of successes in }n \text{ Bernoulli trials}.\\ P\{\mu=k\}=C_n^kp^k(1-p)^{n-k}, k=0,1,\ldots,n.\\ \text{Then }P\{\mu\leq k\}=\sum_{i=0}^kC_n^ip^i(1-p)^{n-i}.\\ \mu\in B(13,0.1).\\ P\{\mu\leq4\}=\sum_{i=0}^4C_{13}^i(0.1)^i(0.9)^{13-i}=(0.9)^{13}+13(0.1)(0.9)^{12}+78(0.1)^2(0.9)^{11}+22\cdot 13(0.1)^3(0.9)^{10}+5\cdot11\cdot13(0.1)^4(0.9)^9\approx0.9935.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS