Answer:
(a) X = 0 (FFFF)
(b) X = 4 (SSSS)
(c) P(x; n, p) = nCx pˣ (1 - p)ⁿ⁻ˣ
X | P(X =x)
0 | 0.2015
1 | 0.3970
2 | 0.2933
3 | 0.0963
4 | 0.0118
(d) E(X) = 1
(e) P(x ≥ 2) = 0.4014
Explanation:
33% of all homes are built to withstand a 9.0 magnitude earthquake.
Let X denote the number among the four that can withstand a level 9.0 magnitude earthquake.
Let S denote a home that can withstand a 9.0 magnitude earthquake
Let F denote a home that can not withstand a 9.0 magnitude earthquake
(a) What is the minimum possible value of X?
The minimum possible value of X is when none of the four selected homes can withstand a level 9.0 magnitude earthquake.
X = 0 (FFFF)
(b) What is the maximum possible value of X?
The maximum possible value of X is when all of the four selected homes can withstand a level 9.0 magnitude earthquake.
X = 4 (SSSS)
(c) Find X for all possible outcomes. (16 of them)
Outcome | X
SSSS | 4
SSSF | 3
SSFS | 3
SSFF | 2
SFSS | 3
SFSF | 2
SFFS | 2
SFFF | 1
FSSS | 3
FSSF | 2
FSFS | 2
FSFF | 1
FFSS | 2
FFSF | 1
FFFS | 1
FFFF | 0
Find the probability distribution of X.
The probability distribution of X can be modeled as a binomial distribution.
We have n = 4 and p = 0.33
P(x; n, p) = nCx pˣ (1 - p)ⁿ⁻ˣ
Where x is the variable of interest, n is the number of trials and p is the probability of success and 1 - p is the probability of failure
For x = 0:
P(x = 0) = ⁴C₀ 0.33⁰ (1 - 0.33)⁴⁻⁰ = 0.2015
For x = 1:
P(x = 1) = ⁴C₁ 0.33¹ (1 - 0.33)⁴⁻¹ = 0.3970
For x = 2:
P(x = 2) = ⁴C₂ 0.33² (1 - 0.33)⁴⁻² = 0.2933
For x = 3:
P(x = 3) = ⁴C₃ 0.33³ (1 - 0.33)⁴⁻³ = 0.0963
For x = 4:
P(x = 4) = ⁴C₄ 0.33⁴ (1 - 0.33)⁴⁻⁴ = 0.0118
X | P(X =x)
0 | 0.2015
1 | 0.3970
2 | 0.2933
3 | 0.0963
4 | 0.0118
(d) What is the most likely value for X?
The most likely value of X is the expected value of X
E(X) = n×p
E(X) = 4×0.33
E(X) = 1.32
Rounding off to nearest whole number
E(X) = 1
(e) What is the probability that at least two of the four selected can withstand a magnitude 9.0 earthquake?
At least two means two or greater than two.
P(x ≥ 2) = P(x = 2) + P(x = 3) + P(x = 4)
P(x ≥ 2) = 0.2933 + 0.0963 + 0.0118
P(x ≥ 2) = 0.4014
Therefore, there is 0.4014 probability that at least two of the four selected can withstand a magnitude 9.0 earthquake.