Question

Some parts of California are particularly earthquake-prone.

Answer 1

Answer:

a) $P(X=0)=(4C0)(0.27)^0 (1-0.27)^{4-0}=0.283982$

$P(X=1)=(4C1)(0.27)^4 (1-0.27)^{4-1}=0.420138$

$P(X=2)=(4C2)(0.27)^5 (1-0.27)^{4-2}=0.23309$

$P(X=3)=(4C3)(0.27)^3 (1-0.27)^{4-3}=0.057474$

$P(X=4)=(4C4)(0.27)^4 (1-0.27)^{4-4}=0.05314$

b) For this case the value with the higher probability is X=1, so then would be the most likely value.

c) $P(X \leq 1) = P(X=0) +P(X=1) = 0.283982+0.420138=0.704121 \approx 0.7041$

d) $P(X \leq 2) =1-{(X$

e) For this case the expected value is given by:

$E(X) = \mu = np= 4*0.27= 1.08$

The variance is given by:

$\sigma^2 = Var(X) = np(1-p) =4*0.27*(1-0.27)= 0.7884$

And the deviation is:

$Sd(X) = \sigma = \sqrt{0.7884}=0.89$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=4, p=0.27)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

$P(X=0)=(4C0)(0.27)^0 (1-0.27)^{4-0}=0.283982$

$P(X=1)=(4C1)(0.27)^4 (1-0.27)^{4-1}=0.420138$

$P(X=2)=(4C2)(0.27)^5 (1-0.27)^{4-2}=0.23309$

$P(X=3)=(4C3)(0.27)^3 (1-0.27)^{4-3}=0.057474$

$P(X=4)=(4C4)(0.27)^4 (1-0.27)^{4-4}=0.05314$

Part b

For this case the value with the higher probability is X=1, so then would be the most likely value.

Part c

For this case we want this probability:

$P(X \leq 1)$

And we can find this probability like this:

$P(X \leq 1) = P(X=0) +P(X=1) = 0.283982+0.420138=0.704121 \approx 0.7041$

Part d

For this case we want this probability:

$P(X \leq 2)$

And we can find this probability using the complement rule like this:

$P(X \leq 2) =1-{(X$

Part e

For this case the expected value is given by:

$E(X) = \mu = np= 4*0.27= 1.08$

The variance is given by:

$\sigma^2 = Var(X) = np(1-p) =4*0.27*(1-0.27)= 0.7884$

And the deviation is:

$Sd(X) = \sigma = \sqrt{0.7884}=0.89$