Question

Problem #1

Some studies suggested that six in 10 people are satisfied with their local bank. If 9 people are randomly selected, what is the probability that the number of satisfied with their local bank is:

1. Exactly two?
2. At most two?
3. At least three?
4. Between four and six, inclusive?

Answer 1

a)

Here, n = 9, p = 0.6, (1 - p) = 0.4 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) = 9C2 * 0.6^2 * 0.4^7
P(X = 2) = 0.0212
0

b)

Here, n = 9, p = 0.6, (1 - p) = 0.4 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) = (9C0 * 0.6^0 * 0.4^9) + (9C1 * 0.6^1 * 0.4^8) + (9C2 * 0.6^2 * 0.4^7)
P(X <= 2) = 0.0003 + 0.0035 + 0.0212
P(X <= 2) = 0.025

c)

Here, n = 9, p = 0.6, (1 - p) = 0.4 and x = 3
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 3).
P(X >= 3) = (9C3 * 0.6^3 * 0.4^6) + (9C4 * 0.6^4 * 0.4^5) + (9C5 * 0.6^5 * 0.4^4) + (9C6 * 0.6^6 * 0.4^3) + (9C7 * 0.6^7 * 0.4^2) + (9C8 * 0.6^8 * 0.4^1) + (9C9 * 0.6^9 * 0.4^0)
P(X >= 3) = 0.0743 + 0.1672 + 0.2508 + 0.2508 + 0.1612 + 0.0605 + 0.0101
P(X >= 3) = 0.9749

d)

Here, n = 9, p = 0.6, (1 - p) = 0.4, x1 = 4 and x2 = 6.
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(4 <= X <= 6)
P(4 <= X <= 6) = (9C4 * 0.6^4 * 0.4^5) + (9C5 * 0.6^5 * 0.4^4) + (9C6 * 0.6^6 * 0.4^3)
P(4 <= X <= 6) = 0.1672 + 0.2508 + 0.2508
P(4 <= X <= 6) = 0.6688