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?

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:

nCx is calculated as:

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:

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

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:

answer is 144

step-by-step explanation:

the answer is (-4,-4)

hope this ! : d