Question

Find the indicated probabilities using the geometric​ distribution, the Poisson​ distribution, or the binomial distribution. Then determine if the events are unusual. If​ convenient, use the appropriate probability table or technology to find the probabilities.

Fifty-nine percent of adults say that they have cheated on a test or exam before. You randomly select six adults. Find the probability that the number of adults who say that they have cheated on a test or exam before is​ (a) exactly four​,​(b) more than two​, and​ (c) at most five.

Answer 1

a)

Here, n = 6, p = 0.59, (1 - p) = 0.41 and x = 4
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 4)
P(X = 4) = 6C4 * 0.59^4 * 0.41^2
P(X = 4) = 0.3055
0

b)

Here, n = 6, p = 0.59, (1 - p) = 0.41 and x = 2
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X > 2).
P(X >2) = (6C3 * 0.59^3 * 0.41^3) + (6C4 * 0.59^4 * 0.41^2) + (6C5 * 0.59^5 * 0.41^1) + (6C6 * 0.59^6 * 0.41^0)
P(X >2) = 0.2831 + 0.3055 + 0.1759 + 0.0422
P(X >2) = 0.8067

c)

Here, n = 6, p = 0.59, (1 - p) = 0.41 and x = 5
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 5).
P(X <= 5) = (6C0 * 0.59^0 * 0.41^6) + (6C1 * 0.59^1 * 0.41^5) + (6C2 * 0.59^2 * 0.41^4) + (6C3 * 0.59^3 * 0.41^3) + (6C4 * 0.59^4 * 0.41^2) + (6C5 * 0.59^5 * 0.41^1)
P(X <= 5) = 0.0048 + 0.041 + 0.1475 + 0.2831 + 0.3055 + 0.1759
P(X <= 5) = 0.9578

All events are not unusual because probability is greater than 0.05