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.
P| 1 2 3 4 5
P(x)| 0.70 0.15 0.10 0.03 0.02
Find P(4)
Find P(greater than 3)
Find the probability that a car only has one occupant.
Find the probability that a car has fewer than 4 occupants.
Compute the mean UvX. Round the answer to 2 decimal places.
Compute the standard deviation OvX. Round to 3 decimal places.
(g) To save energy, a goal is set to have the mean number of occupants be at least two per car. Has this goal been met?
Since the computed mean number of occupants per car is only [blank], then this goal to have the mean number of occupants to be at least two per car [has or has not?] been met.
P| 1 2 3 4 5
P(x)| 0.70 0.15 0.10 0.03 0.02
Find P(4)
Find P(greater than 3)
Find the probability that a car only has one occupant.
Find the probability that a car has fewer than 4 occupants.
Compute the mean UvX. Round the answer to 2 decimal places.
Compute the standard deviation OvX. Round to 3 decimal places.
(g) To save energy, a goal is set to have the mean number of occupants be at least two per car. Has this goal been met?
Since the computed mean number of occupants per car is only [blank], then this goal to have the mean number of occupants to be at least two per car [has or has not?] been met.
Answers
GPT-4o mini
To solve the given problems, we first define the probability distribution of the random variable \(X\), which represents the number of occupants in a randomly chosen car:
\[
\begin{array}{c|ccccc}
X & 1 & 2 & 3 & 4 & 5 \\
\hline
P(X) & 0.70 & 0.15 & 0.10 & 0.03 & 0.02 \\
\end{array}
\]
### (a) Find \( P(4) \)
The probability that a car has 4 occupants is given directly from the probability distribution:
\[
P(4) = 0.03
\]
### (b) Find \( P(X > 3) \)
To find the probability that a car has more than 3 occupants, we sum the probabilities for \(X = 4\) and \(X = 5\):
\[
P(X > 3) = P(4) + P(5) = 0.03 + 0.02 = 0.05
\]
### (c) Find the probability that a car only has one occupant
The probability that a car has only one occupant, \(P(1)\), is given directly:
\[
P(1) = 0.70
\]
### (d) Find the probability that a car has fewer than 4 occupants
To find the probability that a car has fewer than 4 occupants, we sum the probabilities for \(X = 1\), \(X = 2\), and \(X = 3\):
\[
P(X < 4) = P(1) + P(2) + P(3) = 0.70 + 0.15 + 0.10 = 0.95
\]
### (e) Compute the mean \( \mu_X \)
The mean is calculated using the formula:
\[
\mu_X = E[X] = \sum (x \cdot P(X = x))
\]
Calculating this step-by-step:
\[
\mu_X = 1 \cdot 0.70 + 2 \cdot 0.15 + 3 \cdot 0.10 + 4 \cdot 0.03 + 5 \cdot 0.02
\]
\[
= 0.70 + 0.30 + 0.30 + 0.12 + 0.10 = 1.52
\]
So, rounding to two decimal places:
\[
\mu_X \approx 1.52
\]
### (f) Compute the standard deviation \( \sigma_X \)
First, we need to calculate the variance \( \sigma^2_X \):
\[
\sigma^2_X = E[X^2] - (E[X])^2
\]
where
\[
E[X^2] = \sum (x^2 \cdot P(X = x))
\]
Calculating \(E[X^2]\):
\[
E[X^2] = 1^2 \cdot 0.70 + 2^2 \cdot 0.15 + 3^2 \cdot 0.10 + 4^2 \cdot 0.03 + 5^2 \cdot 0.02
\]
\[
= 0.70 + 0.60 + 0.90 + 0.48 + 0.50 = 3.18
\]
Now we can find the variance:
\[
\sigma^2_X = 3.18 - (1.52)^2 = 3.18 - 2.3104 = 0.8696
\]
Now compute the standard deviation:
\[
\sigma_X = \sqrt{0.8696} \approx 0.933
\]
Rounded to three decimal places:
\[
\sigma_X \approx 0.933
\]
### (g) Has the goal been met?
The goal is to have the mean number of occupants be at least two per car. Since the computed mean number of occupants per car is only \(1.52\), then this goal to have the mean number of occupants be at least two per car **has not** been met.
To summarize:
- \( P(4) = 0.03 \)
- \( P(X > 3) = 0.05 \)
- \( P(1) = 0.70 \)
- \( P(X < 4) = 0.95 \)
- Mean \(\mu_X \approx 1.52\)
- Standard deviation \(\sigma_X \approx 0.933\)
The final statement would be:
Since the computed mean number of occupants per car is only **1.52**, then this goal to have the mean number of occupants to be at least two per car **has not** been met.
\[
\begin{array}{c|ccccc}
X & 1 & 2 & 3 & 4 & 5 \\
\hline
P(X) & 0.70 & 0.15 & 0.10 & 0.03 & 0.02 \\
\end{array}
\]
### (a) Find \( P(4) \)
The probability that a car has 4 occupants is given directly from the probability distribution:
\[
P(4) = 0.03
\]
### (b) Find \( P(X > 3) \)
To find the probability that a car has more than 3 occupants, we sum the probabilities for \(X = 4\) and \(X = 5\):
\[
P(X > 3) = P(4) + P(5) = 0.03 + 0.02 = 0.05
\]
### (c) Find the probability that a car only has one occupant
The probability that a car has only one occupant, \(P(1)\), is given directly:
\[
P(1) = 0.70
\]
### (d) Find the probability that a car has fewer than 4 occupants
To find the probability that a car has fewer than 4 occupants, we sum the probabilities for \(X = 1\), \(X = 2\), and \(X = 3\):
\[
P(X < 4) = P(1) + P(2) + P(3) = 0.70 + 0.15 + 0.10 = 0.95
\]
### (e) Compute the mean \( \mu_X \)
The mean is calculated using the formula:
\[
\mu_X = E[X] = \sum (x \cdot P(X = x))
\]
Calculating this step-by-step:
\[
\mu_X = 1 \cdot 0.70 + 2 \cdot 0.15 + 3 \cdot 0.10 + 4 \cdot 0.03 + 5 \cdot 0.02
\]
\[
= 0.70 + 0.30 + 0.30 + 0.12 + 0.10 = 1.52
\]
So, rounding to two decimal places:
\[
\mu_X \approx 1.52
\]
### (f) Compute the standard deviation \( \sigma_X \)
First, we need to calculate the variance \( \sigma^2_X \):
\[
\sigma^2_X = E[X^2] - (E[X])^2
\]
where
\[
E[X^2] = \sum (x^2 \cdot P(X = x))
\]
Calculating \(E[X^2]\):
\[
E[X^2] = 1^2 \cdot 0.70 + 2^2 \cdot 0.15 + 3^2 \cdot 0.10 + 4^2 \cdot 0.03 + 5^2 \cdot 0.02
\]
\[
= 0.70 + 0.60 + 0.90 + 0.48 + 0.50 = 3.18
\]
Now we can find the variance:
\[
\sigma^2_X = 3.18 - (1.52)^2 = 3.18 - 2.3104 = 0.8696
\]
Now compute the standard deviation:
\[
\sigma_X = \sqrt{0.8696} \approx 0.933
\]
Rounded to three decimal places:
\[
\sigma_X \approx 0.933
\]
### (g) Has the goal been met?
The goal is to have the mean number of occupants be at least two per car. Since the computed mean number of occupants per car is only \(1.52\), then this goal to have the mean number of occupants be at least two per car **has not** been met.
To summarize:
- \( P(4) = 0.03 \)
- \( P(X > 3) = 0.05 \)
- \( P(1) = 0.70 \)
- \( P(X < 4) = 0.95 \)
- Mean \(\mu_X \approx 1.52\)
- Standard deviation \(\sigma_X \approx 0.933\)
The final statement would be:
Since the computed mean number of occupants per car is only **1.52**, then this goal to have the mean number of occupants to be at least two per car **has not** been met.