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The probability distribution for the number of adults in groups who reported sleepwalking is provided.

Let x bethe number of sleepwalkers in the group of five.

The requirements are as follows:

1)The variable x is anumerical random variable.

2)The sum of the probabilities is computed as:

$$\begin{gathered} 鈭�P\left(x\right)=0.172+0.363+0.306+0.129+0.027+0.002 \\ =0.999 \end{gathered}$$

Therefore,the sum of the probabilities is approximately equal to 1 with a round of error as 0.001.

3) Each value of P(x) is between 0 and 1.

Thus, all the requirements are satisfied.

The mean for a random variable is computed as:

$$\begin{gathered} 渭=鈭�\left[x脳P\left(x\right)\right] \\ =\left(0脳0.172\right)+\left(1脳0.363\right)+\left(2脳0.306\right)+...+\left(5脳0.002\right) \\ =1.48 \\ 鈮�1.5 \end{gathered}$$

Thus, the mean number of sleepwalkers is 1.5.

The standard deviation of the random variable x is computed as;

$蟽=\sqrt{鈭�\left[x^2脳P\left(x\right)\right]-{渭}^2}$

The calculations are as follows;

$$\begin{gathered} 鈭�\left[x^2路P\left(x\right)\right]=\left(0^2脳0.172\right)+\left(1^2脳0.363\right)+\left(2^2脳0.306\right)+...+\left(5^2脳0.002\right) \\ =3.23 \end{gathered}$$

The standard deviation is given as:

$$\begin{gathered} 蟽=\sqrt{鈭�\left[x^2路P\left(x\right)\right]-{渭}^2} \\ =\sqrt{3.23-{1.5}^2} \\ =0.98 \\ 鈮�1.0 \end{gathered}$$

Thus, the standard deviation of x is 1.0.