A market research team compiled the following discrete probability distribution for families residing in Randolph County.In this distribution, x represents the number of evenings the family dines outside their home during a week. $$\begin{array} { | c | c | }
\hline x & P ( x ) \
\hline 0 & 0.30 \
\hline 1 & 0.50 \
\hline 2 & 0.10 \
\hline 3 & 0.10 \
\hline
\end{array}$$ The standard deviation of x is _______________.
A)1.00
B)2.00
C)0.80
D)0.89
E)1.09

Question

A market research team compiled the following discrete probability distribution for families residing in Randolph County.In this distribution, x represents the number of evenings the family dines outside their home during a week. $$\begin{array} { | c | c | } 
\hline x & P ( x ) \
\hline 0 & 0.30 \
\hline 1 & 0.50 \
\hline 2 & 0.10 \
\hline 3 & 0.10 \
\hline
\end{array}$$ The standard deviation of x is _______________.
A)1.00
B)2.00
C)0.80
D)0.89
E)1.09

Quizplus · Accepted Answer

The standard deviation of a discrete probability distribution is calculated using the formula $\sigma = \sqrt{\sum (x_i - \mu)^2 \cdot P(x_i)}$, where $\mu$ is the mean of the distribution. First, calculate the mean: $\mu = (0)(0.30) + (1)(0.50) + (2)(0.10) + (3)(0.10) = 0 + 0.5 + 0.2 + 0.3 = 1.0$. Then, calculate the variance: $\sigma^2 = (0-1)^2(0.30) + (1-1)^2(0.50) + (2-1)^2(0.10) + (3-1)^2(0.10) = (1)(0.30) + (0)(0.50) + (1)(0.10) + (4)(0.10) = 0.30 + 0 + 0.10 + 0.40 = 0.80$. Finally, the standard deviation $\sigma = \sqrt{0.80} \approx 0.89$.

A Market Research Team Compiled the Following Discrete Probability Distribution xP(x)00.3010.5020.1030.10\begin{array} { | c | c | } \hline x & P ( x ) \\\hline 0 & 0.30 \\\hline 1 & 0.50 \\\hline 2 & 0.10 \\\hline 3 & 0.10 \\\hline\end{array}x0123​P(x)0.300.500.100.10​​

A Market Research Team Compiled the Following Discrete Probability Distribution $\begin{array} { | c | c | } \hline x & P ( x ) \\\hline 0 & 0.30 \\\hline 1 & 0.50 \\\hline 2 & 0.10 \\\hline 3 & 0.10 \\\hline\end{array}$