Answered

The probabilities that a patient will have [tex]$0, 1, 2$[/tex], or [tex]$3$[/tex] medical tests performed on entering a hospital are [tex]$\frac{6}{15}, \frac{5}{15}, \frac{3}{15}$[/tex], and [tex]$\frac{1}{15}$[/tex] respectively. Construct a probability distribution.

\begin{tabular}{|l|l|}
\hline
Medical Tests [tex]$x$[/tex] & [tex]$P(x)$[/tex] (Fractions only) \\
\hline
0 & [tex]$\frac{6}{15}$[/tex] \\
\hline
1 & [tex]$\frac{5}{15}$[/tex] \\
\hline
2 & [tex]$\frac{3}{15}$[/tex] \\
\hline
3 & [tex]$\frac{1}{15}$[/tex] \\
\hline
\end{tabular}

What is the probability of 1 medical test being performed? Simplify all fractions.
[tex]\[ P(x=1) = \frac{1}{3} \][/tex]

What is the probability of less than 2 medical tests being performed?
[tex]\[ P(x\ \textless \ 2) = \frac{11}{15} \][/tex]

What is the probability of between 0 and 1 tests being performed, inclusive?
[tex]\[ P(0 \leq x \leq 1) = \frac{11}{15} \][/tex]

What is the mean and standard deviation? Use decimals and round to the nearest thousandth.
[tex]\[
\begin{array}{l}
\mu = 0.933 \\
\sigma = 0.929
\end{array}
\][/tex]



Answer :

GinnyAnswer
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]