Question

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)?

Explanation:

Step1: Calculate the total number of students

Let the total number of students be \(n\).
We know that \(50\%\) of the students own \(2\) pets. Let the number of students who own \(2\) pets be \(x\).
The number of students who own \(1\) pet \(n_1 = 8\), the number of students who own \(3\) pets \(n_3=3\), and the number of students who own \(4\) pets \(n_4 = 2\).
Since \(P(X = 2)=0.5\), and \(P(X)=\frac{\text{Number of students with }X\text{ pets}}{\text{Total number of students}}\), we know that \(n_2=\frac{n}{2}\).
Also, \(n=n_1 + n_2+n_3 + n_4\). Substituting \(n_1 = 8\), \(n_3=3\), \(n_4 = 2\) and \(n_2=\frac{n}{2}\) into the equation:
\(n=8+\frac{n}{2}+3 + 2\)
\(n-\frac{n}{2}=8 + 3+2\)
\(\frac{n}{2}=13\)
\(n = 26\)

Step2: Calculate the probabilities

• For \(X = 1\): \(P(X = 1)=\frac{8}{26}\approx0.3\)
• For \(X = 2\): \(P(X = 2)=\frac{13}{26}=0.5\)
• For \(X = 3\): \(P(X = 3)=\frac{3}{26}\approx0.1\)
• For \(X = 4\): \(P(X = 4)=\frac{2}{26}\approx0.08\approx0.1\)

Answer:

The first probability - distribution graph (where \(P(X = 1)\approx0.3\), \(P(X = 2)=0.5\), \(P(X = 3)\approx0.1\), \(P(X = 4)\approx0.1\)) is the correct probability distribution \(P_X(x)\)