Question

8) Assume that 15?% of people are? left-handed. Suppose 14 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.

b)what's the probability that they're not all right-handed?

c)what's the probability that there are no more than 8 righties?

d)what's the probability that there are exactly 6 of each?

Answer 1

Solution Given The probability ofleft-handed people is 15% or 0.15 Compute the probability ofright handed people P(right-handed people) 1-P(left-handed people) p 1-0.15 =0.85 A random sample of 14 people is selected Let Xrepresent the number of right handed people X~Bin (n 14,p = 0.85) The probability mass function of a binomial distribution is, Р(х - х) — "С,p" (1-р)** The mean and standard deviation of a binomial distribution is Е(X) %3 пр SD.(X)-Vпp(1-р) (a) Compute the mean number ofright handed people in the group of 14 E(X=np -14x0.85 - 119 12 Compute the standard deviation ofthe number ofright handed people in the group of 14

SD.(X)np(1-p) 14x0.85x(1-0.85) =1.785 =1.336 Therefore, the mean and standard deviation of the number of right handed people in the group of 14 is 12 and 1.336 respectively (b) right handed Compute the probability that none of the 14 people are P(X 0)C (0.85) (1-0.85)" 14-0 =1x1x2.92x10-2 -2.92x10-12 right handed is 2.92 x10-12| Therefore, the probability that none of the 14 people are (c) Compute the probability that there are 8 or less right-handed people- Р(X<8)-1-Р(Х >8) 1-P(X9)+P(X-10)+P(X11)+P (X = 12)+ P(X13)+ P(X=14)] =1-C(0.85)(1-0.85)' + C(0.85)(1-0.85).C(0.85) (1-0.8s =1-[0.0352+0.0998+...+0.1028] =1-0.9885 =0.0115 Therefore, the probability that there are 8 or less right-handed people in a group of 14 is 0.0115 (d) There are total 14 people in the sample. It is not possible that the sample consists of exactly 6 of each. Because then there will still be two more people left who are either left handed or right handed