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.

h who knows

step-by-step explanation:

stew reads at the fastest rate of 265 words per minute.

step-by-step explanation:

it's a given that felicia reads at 260 word per minute and on connie's graph we can see that she reads 200 words per minute. since stew's table tells us that he read 795 in 3 minutes, we can divide 795 by 3 to find out how many words he reads per minute.

795/3 = 265

stew reads 265 words per minute - he reads at the fastest rate since felecia reads at 260 words per minute and connie reads at 200 words per minute.

stew (265) > felecia (260) > connie (200)