Question

Assume that 80% 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

Solution:-

a) The probability that there are some lefties (≥ 1) among the 5 people is 0.99968.

x = 1, n = 5, p = 0.80

By applying binomial distribution:-

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P(x > 1) = 0.99968

b) The probability that there are exactly 3 lefties in the group is 0.2048.

x = 3, n = 5, p = 0.80

By applying binomial distribution:-

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P(x = 3) = 0.2048

c) The probability that there are at least 4 lefties in the group is 0.7373.

x = 4, n = 5, p = 0.80

By applying binomial distribution:-

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P(x > 4) = 0.7373.

d) The probability that there are no more than 2 lefties in the group is 0.05792.

x = 2, n = 5, p = 0.80

By applying binomial distribution:-

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P(x < 2) = 0.05792.

e)The expected number of lefties in the group are 4.0.

E(x) = n × p

E(x) = 5 × 0.80

E(x) = 4.0

f) The standard deviation of the lefties in the group is 0.8944.