taylor surveys students in one grade level who own at least one pet. she finds that 50% of the students…

taylor surveys students in one grade level who own at least one pet. she finds that 50% of the students surveyed own 2 pets, 3 students own 3 pets each, and 2 students own 4 pets each. eight of the students in the grade own 1 pet. considering the number of pets as the random variable, x, which of the following is the probability distribution, $p_x(x)$?

taylor surveys students in one grade level who own at least one pet. she finds that 50% of the students surveyed own 2 pets, 3 students own 3 pets each, and 2 students own 4 pets each. eight of the students in the grade own 1 pet. considering the number of pets as the random variable, x, which of the following is the probability distribution, $p_x(x)$?

Answer

Explanation:

Step1: Let the total number of students be $n$.

We know that the proportion of students with 2 - pets is 50% or 0.5. The number of students with 3 - pets is 3, the number of students with 4 - pets is 2, and the number of students with 1 - pet is 8. So, the number of students with 2 - pets is $0.5n$. Then $n=8 + 3+2 + 0.5n$.

Step2: Solve the equation for $n$.

Subtract $0.5n$ from both sides: $n-0.5n=8 + 3+2$. So, $0.5n=13$, and $n = 26$.

Step3: Calculate the probabilities for each value of $X$.

For $X = 1$: $P(X = 1)=\frac{8}{26}=\frac{4}{13}$. For $X = 2$: Since the number of students with 2 - pets is $0.5n=13$, $P(X = 2)=\frac{13}{26}=\frac{1}{2}$. For $X = 3$: $P(X = 3)=\frac{3}{26}$. For $X = 4$: $P(X = 4)=\frac{2}{26}=\frac{1}{13}$.

Answer:

$P(X = 1)=\frac{4}{13},P(X = 2)=\frac{1}{2},P(X = 3)=\frac{3}{26},P(X = 4)=\frac{1}{13}$