Question

refer to the accompanying table, which describes the number of adults in groups of five who reported sleepwalking. find the mean and standard deviation for the numbers of sleepwalkers in groups of five. click the icon to view the data table. the mean is sleepwalker(s) (round to one decimal place as needed.) table of numbers of sleepwalkers and probabilities x p(x) 0 0.194 1 0.355 2 0.242 3 0.129 4 0.028 5 0.002

Explanation:

Step1: Recall mean formula for discrete - probability distribution

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

Step2: Calculate the products and sum them

For $x = 0$, $x\cdot P(x)=0\times0.194 = 0$;
For $x = 1$, $x\cdot P(x)=1\times0.365 = 0.365$;
For $x = 2$, $x\cdot P(x)=2\times0.242 = 0.484$;
For $x = 3$, $x\cdot P(x)=3\times0.129 = 0.387$;
For $x = 4$, $x\cdot P(x)=4\times0.028 = 0.112$;
For $x = 5$, $x\cdot P(x)=5\times0.002 = 0.01$.
Then $\mu=0 + 0.365+0.484 + 0.387+0.112+0.01=\sum_{x = 0}^{5}x\cdot P(x)=1.368\approx1.4$.

Step3: Recall variance formula for discrete - probability distribution

The variance $\sigma^{2}=\sum(x-\mu)^{2}\cdot P(x)$.
First, calculate $(x - \mu)^{2}\cdot P(x)$ for each $x$:
For $x = 0$, $(0 - 1.368)^{2}\times0.194=( - 1.368)^{2}\times0.194 = 1.871424\times0.194\approx0.363$;
For $x = 1$, $(1 - 1.368)^{2}\times0.365=( - 0.368)^{2}\times0.365 = 0.135424\times0.365\approx0.049$;
For $x = 2$, $(2 - 1.368)^{2}\times0.242=(0.632)^{2}\times0.242 = 0.399424\times0.242\approx0.097$;
For $x = 3$, $(3 - 1.368)^{2}\times0.129=(1.632)^{2}\times0.129 = 2.663424\times0.129\approx0.344$;
For $x = 4$, $(4 - 1.368)^{2}\times0.028=(2.632)^{2}\times0.028 = 6.927424\times0.028\approx0.194$;
For $x = 5$, $(5 - 1.368)^{2}\times0.002=(3.632)^{2}\times0.002 = 13.191424\times0.002\approx0.026$.
$\sigma^{2}=0.363 + 0.049+0.097+0.344+0.194+0.026 = 1.073$.

Step4: Calculate the standard deviation

The standard deviation $\sigma=\sqrt{\sigma^{2}}=\sqrt{1.073}\approx1.0$.

Answer:

The mean is $1.4$ sleepwalkers. The standard - deviation is $1.0$ sleepwalkers.