Question

Assume that 63% of people are left-handed. If we select 5 people at random, find the probability of each outcome described below, rounded to four decimal places: a. There are some lefties ( ≥ 1) among the 5 people. b. There are exactly 3 lefties in the group. c. There are at least 4 lefties in the group. d. There are no more than 2 lefties in the group. e. How many lefties do you expect? f. With what standard deviation?

Answer 1

Answer:

a. 0.9931

b. 0.3423

c. 0.3907

d. 0.2670

e. 3.15

f. 1.0796

Step-by-step explanation:

The probability of the variable that said the number of left-handed people follow a Binomial distribution, so the probability is:

$P(x)=nCx*p^{x}*(1-p)^{n-x}$

nCx is calculated as:

$nCx=\frac{n!}{x!(n-x)!}$

Where x is the number of left-handed people, n is the number of people selected at random and p is the probability that a person is left--handed. So P(x) is:

$P(x)=5Cx*0.63^{x}*(1-0.63)^{5-x}$

Then the probabilities P(0), P(1), P(2), P(3), P(4) and P(5) are:

$P(0)=5C0*0.63^{0}*(1-0.63)^{5-0}=0.0069$

$P(1)=5C1*0.63^{1}*(1-0.63)^{5-1}=0.0590$

$P(2)=5C2*0.63^{2}*(1-0.63)^{5-2}=0.2011$

$P(3)=5C3*0.63^{3}*(1-0.63)^{5-3}=0.3423$

$P(x)=5C4*0.63^{4}*(1-0.63)^{5-4}=0.2914$

$P(x)=5C5*0.63^{5}*(1-0.63)^{5-5}=0.0993$

Then, the probability P(x≥1) that there are some lefties among the 5 people is:

P(x≥1) = P(1) + P(2) + P(3) + P(4) + P(5)

P(x≥1) = 0.0590 + 0.2011 + 0.3423 + 0.2914 + 0.0993 = 0.9931

The probability P(3) that there are exactly 3 lefties in the group is:

P(3) = 0.3423

The probability P(x≥4) that there are at least 4 lefties in the group is:

P(x≥4) = P(4) + P(5) = 0.2914 + 0.0993 = 0.3907

The probability P(x≤2) that there are no more than 2 lefties in the group is:

P(x≤2) = P(0) + P(1) + P(2) = 0.0069 + 0.0590 + 0.2011 = 0.2670

On the other hand, the expected value E(x) and standard deviation S(x) of the variable that follows a binomial distribution is:

$E(x)=np=5(0.63)=3.15\\S(x)=\sqrt{np(1-p)}=\sqrt{5(0.63)(1-0.63)}=1.0796$