Assume that 15% of people are left-handed. Suppose 8 people are selected at rand

Question

Assume that 15% of people are left-handed. Suppose 8 people are selected at random. Answer each question about right-handers below. a) Find the mean and standard deviation of the number of right-handers in the group. right-handers right-handers (Round to two decimal places as needed ) b) What's the probability that they're not all right-handed? (Round to three decimal places as needed) c) What's the probability that there are no more than 7 righties? (Round to three decimal places as needed) d) What's the probability that there are exactly 4 of each? (Round to five decimal places as needed.) e) What's the probability that the majority is right-handed? (Round to three decimal places as needed.)

Explanation / Answer

Ans:

Binomial distribution with n=8,p=0.15

P(x=r)=8Cr*0.15r*(1-0.15)8-r

a)Mean=np=8*0.15=1.2

standard deviation=sqrt(np(1-p))=sqrt(8*0.15*(1-0.15))=1.01

b)P(none is right handed)=(1-0.15)8=0.858=0.272

c)P(x<=7)=1-P(x=8)=1-0.158=1-0=1

d)P(x=4)=8C4*0.154*0.854=0.018

e)P(x>4)=1-P(x<=4)=1-BINOMDIST(4,8,0.15,TRUE)=1-0.997=0.003

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