To construct the probability distribution, we fill in the table with the given probabilities:
[tex]\[
\begin{array}{|l|l|}
\hline
\text{Medical Tests } x & P(x) \text{ (Fractions only)} \\
\hline
0 & \frac{6}{15} \\
\hline
1 & \frac{5}{15} \\
\hline
2 & \frac{3}{15} \\
\hline
3 & \frac{1}{15} \\
\hline
\end{array}
\][/tex]
Now, let's answer the specific questions:
1. What is the probability of 1 medical test being performed? Simplify all fractions.
[tex]\[
P(x=1) = \frac{5}{15} = \frac{1}{3}
\][/tex]
2. What is the probability of less than 2 medical tests being performed?
[tex]\[
P(x < 2) = P(x=0) + P(x=1) = \frac{6}{15} + \frac{5}{15} = \frac{11}{15}
\][/tex]
3. What is the probability of between 0 and 1 tests being performed, inclusive?
[tex]\[
P(0 \leq x \leq 1) = P(x=0) + P(x=1) = \frac{6}{15} + \frac{5}{15} = \frac{11}{15}
\][/tex]
4. What are the mean and standard deviation? Use decimals and round to the nearest thousandth.
- The mean (μ) of the distribution is:
[tex]\[
\mu = 0 \cdot \frac{6}{15} + 1 \cdot \frac{5}{15} + 2 \cdot \frac{3}{15} + 3 \cdot \frac{1}{15} = 0.933
\][/tex]
- The variance (σ²) is:
[tex]\[
\begin{align*}
\sigma^2 &= \left(0 - \mu\right)^2 \cdot \frac{6}{15} + \left(1 - \mu\right)^2 \cdot \frac{5}{15} + \left(2 - \mu\right)^2 \cdot \frac{3}{15} + \left(3 - \mu\right)^2 \cdot \frac{1}{15} \\
\end{align*}
\][/tex]
- The standard deviation (σ) is the square root of the variance:
[tex]\[
\sigma = 0.929
\][/tex]
So, summarizing:
[tex]\[
\begin{array}{l}
\mu = 0.933 \\
\sigma = 0.929
\end{array}
\][/tex]
