Let's find the mean and standard deviation for the given distribution of the number of sleepwalkers in groups of five.
### Mean (Expected Value)
The mean [tex]\( \mu \)[/tex] of a probability distribution [tex]\( P(x) \)[/tex] can be calculated using the formula:
[tex]\[ \mu = \sum_{x} [x \cdot P(x)] \][/tex]
Where [tex]\( x \)[/tex] is the possible value of the random variable and [tex]\( P(x) \)[/tex] is the probability of [tex]\( x \)[/tex].
Given the table:
[tex]\[
\begin{array}{c|c}
x & P(x) \\
\hline
0 & 0.182 \\
1 & 0.358 \\
2 & 0.311 \\
3 & 0.119 \\
4 & 0.029 \\
5 & 0.001 \\
\end{array}
\][/tex]
We can compute the mean as follows:
[tex]\[
\mu = (0 \times 0.182) + (1 \times 0.358) + (2 \times 0.311) + (3 \times 0.119) + (4 \times 0.029) + (5 \times 0.001)
\][/tex]
[tex]\[
\mu = 0 + 0.358 + 0.622 + 0.357 + 0.116 + 0.005
\][/tex]
[tex]\[
\mu = 1.458
\][/tex]
### Variance
Variance [tex]\( \sigma^2 \)[/tex] is calculated as:
[tex]\[ \sigma^2 = \sum_{x} [(x - \mu)^2 \cdot P(x)] \][/tex]
Using the values of [tex]\( x \)[/tex], [tex]\( P(x) \)[/tex], and [tex]\( \mu = 1.458 \)[/tex]:
[tex]\[
\sigma^2 = (0 - 1.458)^2 \cdot 0.182 + (1 - 1.458)^2 \cdot 0.358 + (2 - 1.458)^2 \cdot 0.311 + (3 - 1.458)^2 \cdot 0.119 + (4 - 1.458)^2 \cdot 0.029 + (5 - 1.458)^2 \cdot 0.001
\][/tex]
### Standard Deviation
The standard deviation [tex]\( \sigma \)[/tex] is the square root of the variance:
[tex]\[ \sigma = \sqrt{\sigma^2} \][/tex]
Given the numerical results:
[tex]\[
\mu = 1.458
\][/tex]
[tex]\[
\sigma \approx 1.018
\][/tex]
So, the mean number of sleepwalkers in groups of five is 1.458, and the standard deviation is approximately 1.018.
