Question

Let X represent the number of occupants in a randomly chosen car on a certain stretch of highway during morning commute hours. A survey of cars showed that the probability distribution of X is as follows.

X| 1 2 3 4 5
P(X)| 0.65 0.13 0.10 0.03 0.09

(a) Find P(5)

which should be 0.09?

.
Answer
Find P (less than or equal to 3)
Answer
Find the probability the car only has one occupant
Answer
Find the probability that a car has fewer than 4 occupants
Answer
Compute the mean UvX
Answer
Compute the standard deviation OvX. Round the answer to 3 decimal places
Answered by GPT-4o mini
Yes, that's correct! According to the provided probability distribution of \( X \):

- \( P(X=5) = 0.09 \)

So when you asked for \( P(5) \), it is indeed equal to **0.09**.

If you have any other questions about the probability distribution or any other related topic, feel free to ask!
Answered by GPT-4o mini
To find \( P(X \leq 3) \), you need to sum the probabilities of all outcomes where \( X \) is less than or equal to 3. According to your probability distribution:

- \( P(X=1) = 0.65 \)
- \( P(X=2) = 0.13 \)
- \( P(X=3) = 0.10 \)

Now, calculate \( P(X \leq 3) \):

\[
P(X \leq 3) = P(X=1) + P(X=2) + P(X=3)
\]

Substituting the values:

\[
P(X \leq 3) = 0.65 + 0.13 + 0.10 = 0.88
\]

So, \( P(X \leq 3) = 0.88 \).
Answered by GPT-4o mini
The probability that a randomly chosen car has only one occupant, represented by \( P(X = 1) \), can be directly taken from the probability distribution you provided.

From the distribution, we have:

- \( P(X = 1) = 0.65 \)

Therefore, the probability that a car has only one occupant is **0.65**.
Answered by GPT-4o mini
To find the probability that a car has fewer than 4 occupants, we need to calculate \( P(X < 4) \), which is the same as \( P(X = 1) + P(X = 2) + P(X = 3) \).

From your probability distribution, we have:

- \( P(X=1) = 0.65 \)
- \( P(X=2) = 0.13 \)
- \( P(X=3) = 0.10 \)

Now, sum these probabilities:

\[
P(X < 4) = P(X=1) + P(X=2) + P(X=3)
\]

Substituting the values:

\[
P(X < 4) = 0.65 + 0.13 + 0.10 = 0.88
\]

Therefore, the probability that a car has fewer than 4 occupants is **0.88**.
Answered by GPT-4o mini
To compute the mean (or expected value) of the random variable \( X \), denoted as \( E(X) \) or \( \mu_X \), you can use the following formula:

\[
E(X) = \sum_{i=1}^{n} x_i P(X = x_i)
\]

In this case, the values of \( X \) and the corresponding probabilities are as follows:

- \( X = 1, P(X=1) = 0.65 \)
- \( X = 2, P(X=2) = 0.13 \)
- \( X = 3, P(X=3) = 0.10 \)
- \( X = 4, P(X=4) = 0.03 \)
- \( X = 5, P(X=5) = 0.09 \)

Now, we can calculate \( E(X) \):

\[
E(X) = (1)(0.65) + (2)(0.13) + (3)(0.10) + (4)(0.03) + (5)(0.09)
\]

Calculating each term:

\[
E(X) = 0.65 + 0.26 + 0.30 + 0.12 + 0.45
\]

Now, sum these values:

\[
E(X) = 0.65 + 0.26 + 0.30 + 0.12 + 0.45 = 1.78
\]

Therefore, the mean \( E(X) = 1.78 \).
Answered by GPT-4o mini
To compute the standard deviation of the random variable \( X \), we first need to find the variance (\( Var(X) \)) using the following formula:

\[
Var(X) = E(X^2) - (E(X))^2
\]

We already calculated \( E(X) \) (the mean) to be \( 1.78 \). Next, we need to calculate \( E(X^2) \):

\[
E(X^2) = \sum_{i=1}^{n} x_i^2 P(X = x_i)
\]

Using the probabilities you provided:

- \( X = 1, P(X=1) = 0.65 \)
- \( X = 2, P(X=2) = 0.13 \)
- \( X = 3, P(X=3) = 0.10 \)
- \( X = 4, P(X=4) = 0.03 \)
- \( X = 5, P(X=5) = 0.09 \)

Now, let's compute \( E(X^2) \):

\[
E(X^2) = (1^2)(0.65) + (2^2)(0.13) + (3^2)(0.10) + (4^2)(0.03) + (5^2)(0.09)
\]

Calculating each term:

\[
E(X^2) = (1)(0.65) + (4)(0.13) + (9)(0.10) + (16)(0.03) + (25)(0.09)
\]

\[
E(X^2) = 0.65 + 0.52 + 0.90 + 0.48 + 2.25
\]

Now sum these values:

\[
E(X^2) = 0.65 + 0.52 + 0.90 + 0.48 + 2.25 = 5.80
\]

Now we can calculate the variance:

\[
Var(X) = E(X^2) - (E(X))^2
\]
\[
Var(X) = 5.80 - (1.78)^2
\]
\[
Var(X) = 5.80 - 3.1684 = 2.6316
\]

Finally, the standard deviation (\( \sigma_X \)) is the square root of the variance:

\[
\sigma_X = \sqrt{Var(X)} = \sqrt{2.6316} \approx 1.620
\]

Rounded to three decimal places, the standard deviation \( O_v X \) is:

\[
\boxed{1.620}
\]