Compute the Variance and Standard Deviation of a Discrete Random Variable
-The random variable $x$ represents the number of tests that a patient entering a clinic will have along with the corresponding probabilities. Find the mean and standard deviation for the random variable $x$.
$\begin{array}{l|c}
\mathrm{x} & \mathrm{P}(\mathrm{x}) \
\hline 0 & \frac{3}{17} \
1 & \frac{5}{17} \
2 & \frac{6}{17} \
3 & \frac{2}{17} \
4 & \frac{1}{17}
\end{array}$

A) mean: 1.59; standard deviation: $1.09$
B) mean: $1.59$; standard deviation: $3.72$
C) mean: $2.52$; standard deviation: $1.93$
D) mean: $3.72$; standard deviation: $2.52$

Question

Compute the Variance and Standard Deviation of a Discrete Random Variable
-The random variable $x$ represents the number of tests that a patient entering a clinic will have along with the corresponding probabilities. Find the mean and standard deviation for the random variable $x$.
$\begin{array}{l|c}
\mathrm{x} & \mathrm{P}(\mathrm{x}) \
\hline 0 & \frac{3}{17} \
1 & \frac{5}{17} \
2 & \frac{6}{17} \
3 & \frac{2}{17} \
4 & \frac{1}{17}
\end{array}$

A) mean: 1.59; standard deviation: $1.09$
B) mean: $1.59$; standard deviation: $3.72$
C) mean: $2.52$; standard deviation: $1.93$
D) mean: $3.72$; standard deviation: $2.52$

Quizplus · Accepted Answer

To find the mean ($\mu$) of the random variable $x$, we multiply each value of $x$ by its probability and sum these products: $\mu = (0)\left(\frac{3}{17}\right) + (1)\left(\frac{5}{17}\right) + (2)\left(\frac{6}{17}\right) + (3)\left(\frac{2}{17}\right) + (4)\left(\frac{1}{17}\right) = 1.59$. To find the variance ($\sigma^2$), we calculate $\sigma^2 = \sum [x^2 \cdot P(x)] - \mu^2$, and then take the square root of the variance to find the standard deviation ($\sigma$). The calculated standard deviation is approximately $1.09$.

Compute the Variance and Standard Deviation of a Discrete Random xxx

Compute the Variance and Standard Deviation of a Discrete Random $x$