The random variable x is the number of houses sold by a realtor in a single month at the Sendsom's Real Estate Office.Its probability distribution is as follows.Find the standard deviation of the number of houses sold. $$\begin{array} { c | c }
\text { Houses Sold } ( \mathrm { x } ) & \text { Probability P(x) } \
\hline 0 & 0.24 \
\hline 1 & 0.01 \
\hline 2 & 0.12 \
\hline 3 & 0.16 \
\hline 4 & 0.01 \
\hline 5 & 0.14 \
\hline 6 & 0.11 \
\hline 7 & 0.21
\end{array}$$
A)6.86
B)2.62
C)1.62
D)2.25
E)4.45

Question

The random variable x is the number of houses sold by a realtor in a single month at the Sendsom's Real Estate Office.Its probability distribution is as follows.Find the standard deviation of the number of houses sold. $$\begin{array} { c | c } 
\text { Houses Sold } ( \mathrm { x } ) & \text { Probability P(x) } \
\hline 0 & 0.24 \
\hline 1 & 0.01 \
\hline 2 & 0.12 \
\hline 3 & 0.16 \
\hline 4 & 0.01 \
\hline 5 & 0.14 \
\hline 6 & 0.11 \
\hline 7 & 0.21
\end{array}$$
A)6.86
B)2.62
C)1.62
D)2.25
E)4.45

Quizplus · Accepted Answer

To find the standard deviation of the number of houses sold, we first need to calculate the mean ($\mu$) and then use it to find the variance ($\sigma^2$). The standard deviation ($\sigma$) is the square root of the variance.1. Calculate the mean ($\mu$):$\mu = \sum [x \cdot P(x)] = (0 \cdot 0.24) + (1 \cdot 0.01) + (2 \cdot 0.12) + (3 \cdot 0.16) + (4 \cdot 0.01) + (5 \cdot 0.14) + (6 \cdot 0.11) + (7 \cdot 0.21) = 3.59$2. Calculate the variance ($\sigma^2$):$\sigma^2 = \sum [(x - \mu)^2 \cdot P(x)]$$\sigma^2 = (0-3.59)^2 \cdot 0.24 + (1-3.59)^2 \cdot 0.01 + (2-3.59)^2 \cdot 0.12 + (3-3.59)^2 \cdot 0.16 + (4-3.59)^2 \cdot 0.01 + (5-3.59)^2 \cdot 0.14 + (6-3.59)^2 \cdot 0.11 + (7-3.59)^2 \cdot 0.21$$\sigma^2 = 6.8641$3. Calculate the standard deviation ($\sigma$):$\sigma = \sqrt{\sigma^2} = \sqrt{6.8641} = 2.62$Therefore, the standard deviation of the number of houses sold is 2.62.

The Random Variable X Is the Number of Houses Sold