Question

find the mean and standard deviation for the numbers of sleepwalkers in groups of five. table of numbers of sleepwalkers and probabilities x p(x) 0 0.167 1 0.365 2 0.298 3 0.139 4 0.028 5 0.003

Explanation:

Step1: Recall mean formula

The mean $\mu$ of a discrete - probability distribution is given by $\mu=\sum_{x}x\cdot P(x)$.
\[

$$\begin{align*} \mu&=(0\times0.167)+(1\times0.365)+(2\times0.298)+(3\times0.139)+(4\times0.028)+(5\times0.003)\\ &=0 + 0.365+0.596 + 0.417+0.112 + 0.015\\ &=1.505 \end{align*}$$

\]

Step2: Recall variance formula

The variance $\sigma^{2}=\sum_{x}(x - \mu)^{2}\cdot P(x)$.
\[

$$\begin{align*} &(0 - 1.505)^{2}\times0.167+(1 - 1.505)^{2}\times0.365+(2 - 1.505)^{2}\times0.298+(3 - 1.505)^{2}\times0.139+(4 - 1.505)^{2}\times0.028+(5 - 1.505)^{2}\times0.003\\ &=( - 1.505)^{2}\times0.167+( - 0.505)^{2}\times0.365+(0.495)^{2}\times0.298+(1.495)^{2}\times0.139+(2.495)^{2}\times0.028+(3.495)^{2}\times0.003\\ &=2.265025\times0.167 + 0.255025\times0.365+0.245025\times0.298+2.235025\times0.139+6.225025\times0.028+12.215025\times0.003\\ &=0.378269+0.093084+0.073018+0.310669+0.174301+0.036645\\ &=1.065986 \end{align*}$$

\]

Step3: Calculate standard deviation

The standard deviation $\sigma=\sqrt{\sigma^{2}}$. So $\sigma=\sqrt{1.065986}\approx1.032$.

Answer:

The mean is $1.505$ and the standard deviation is approximately $1.032$.