can anyone solves it?please. HW4 - Discrete Distributions: Problem 4 Previous Pr

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can anyone solves it?please.

HW4 - Discrete Distributions: Problem 4 Previous Problem List Next (1 point) Suppose the number of cell phones in a household has a binomial distribution with parameters n Find the probablity of a household having: (a) 3 or 5 cell phones (b) 3 or fewer cell phones (c) 12 or more cell phones (d) fewer than 5 cell phones (e) more than 3 cell phones -21 and p 85% Note: You can earn partial credit on this problem Preview My Answers Submit Answers You have attempted this problem 1 time. Your overall recorded score is 096. You have unlimited attempts remaining.

Explanation / Answer

Ans:

Binomial distribution with n=21 and p=0.85

P(x=r)=21Cr*0.85r*(1-0.85)21-r

a)P(3 or 5)=P(x=3)+Px=5)=0.0000+0.0000=0.0000

b)P(x<=3)=0.0000

c)P(x>=12)=1-P(x<=11)=1-0.0001-0.0003=0.9996

d)P(x<5)=0.0000

e)P(x>3)=1-P(x<=3)=1-0.0000=1.0000

x p(x) 0 0.0000 1 0.0000 2 0.0000 3 0.0000 4 0.0000 5 0.0000 6 0.0000 7 0.0000 8 0.0000 9 0.0000 10 0.0001 11 0.0003 12 0.0016 13 0.0063 14 0.0204 15 0.0540 16 0.1147 17 0.1912 18 0.2408 19 0.2155 20 0.1221 21 0.0329

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