Twenty percent of consumers prefer to purchase groceries online. You randomly select 16 consumers. Find the probability that the number of consumers who prefer to purchase groceries online is (a) exactly one, (b) more than one, and (c) at most one.

a) 0.113

b) 0.859

c) 0.141

Step-by-step explanation:

We would assume a binomial distribution for the number of consumers who prefer to purchase groceries online.. The formula is expressed as

P(x = r) = nCr × p^r × q^(n - r)

Where

x represent the number of successes.

p represents the probability of success.

q = (1 - r) represents the probability of failure.

n represents the number of trials or sample.

From the information given,

p = 20% = 20/100 = 0.2

q = 1 - p = 1 - 0.2

q = 0.8

n = 16

a) P(x = 1)

P(x = 1) = 16C1 × 0.2^1 × 0.8^(16 - 1)

P(x = 1) = 0.113

b) P(x > ) = 1 - P(x ≤ 1)

P(x ≤ 1) = P(x = 0) + P(x = 1)

P(x = 0) = 16C0 × 0.2^0 × 0.8^(16 - 0

P(x = 0) = 0.028

P(x = 1) = 0.113

P(x > ) = 1 - (0.028 + 0.113) = 0.859

c) P(≤ 1) = P(x = 0) + P(x = 1)

P(≤ 1) = 0.028 + 0.113 = 0.141

its a i did it myself

step-by-step explanation: