Assume that when adults with smartphones are randomly selected, 54% use them in meetings or classes. If 12 adult smartphone users are randomly selected, find the probability that fewer than 3 of them use their smartphones in meetings or classes. Round to 5 decimal places.

Answer: P (x < 3) = 0.0096

Step-by-step explanation:

We would assume a binomial distribution for the number of adults that use their smart phones in meetings or classes. 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 = 54% = 54/100 = 0.54

q = 1 - p = 1 - 0.54

q = 0.46

n = 12

P (x < 3) = P (x = 0) + P (x = 1) + P (x = 2)

Therefore,

P (x = 0) = 12C0 * 0.54^0 * 0.46^ (12 - 0) = 0.0000897

P (x = 1) = 12C1 * 0.54^1 * 0.46^ (12 - 1) = 0.0013

P (x = 2) = 12C2 * 0.54^2 * 0.46^ (12 - 2) = 0.0082

Therefore,

P (x < 3) = 0.0000897 + 0.0013 + 0.0082 = 0.0096