Some parts of California are particularly earthquake-prone. Suppose that in one such area, 33% of all homes are built to withstand a 9.0 magnitude earthquake. Four homes are to be selected at random. Let Xdenote 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 and F one that can not. Then one possible outcome is SFSS, with probability (0.33)(0.67)(0.33)(0.33) and associated X value 3. (a) What is the minimum possible value of X?
(b) What is the maximum possible value of X?
(c) Find X for all possible outcomes. (16 of them)
Find the probability distribution of X.
(d) What is the most likely value for X?
(e) What is the probability that at least two of the four selected can withstand a magnitude 9.0 earthquake?

(a) X = 0 ()

(b) X = 4 ()

(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

Step-by-step 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 ()

(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 ()

(c) Find X for all possible outcomes. (16 of them)

Outcome  |    X

       |    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

       |    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.

option a.

step-by-step explanation:

if you look closely at the graph, we can notice that the function is not defined at two different points.

if you look at functions b and c, you will notice that both functions are not defined at just one point. the function b is not defined at x= -5/2 and function c is not defined at x=-5. both options are discarded, given that our function should not be defined at two fiferent points.

now, if you look closely at the graph you will notice that the function sketched is not defined at x=5 and x=-2.