91影视

a.

From the information, it is clear that a game involves large wheel with 20 nickel values.

That is,$\left(5,10,15,20,25,30,35,40,35,50,55,60,65,70,75,80,85,90,95,100\right)$

Let x represents the total score for a single contestant playing.

Since, the 20 possible outcomes are equally likely, the probability of choosing any of the outcomes is clearly $\frac{1}{20}$.

The probability distribution of x is obtained below:

That is,

$\begin{array}{l}P\left(x=5\right)=\frac{1}{20}\\ P\left(x=10\right)=\frac{1}{20}\end{array}$

Similarly for the remaining values are shown in the probability distribution table.

Hence p(x) =0.05 for all the values of 鈥榵鈥�.

The formula for the mean is given below

$\begin{array}{l}\mathbf{渭}\mathbf{=}\mathbf{E}\left(x\right)\\ \mathbf{}\mathbf{}\mathbf{=}\underset{\mathbf{}}{\mathbf{鈭�}}\mathbf{x}\mathbf{}\mathbf{p}\left(x\right)\end{array}$

$$\begin{gathered} \left\{\begin{array}{c}5\frac{1}{20}+10\left(\frac{1}{20}\right)+15\left(\frac{1}{20}\right)+20\left(\frac{1}{20}\right)+25\left(\frac{1}{20}\right)\\ +30\left(\frac{1}{20}\right)+35\left(\frac{1}{20}\right)+40\left(\frac{1}{20}\right)+45\left(\frac{1}{20}\right)+50\left(\frac{1}{20}\right)\\ +55\left(\frac{1}{20}\right)+60\left(\frac{1}{20}\right)+65\left(\frac{1}{20}\right)+70\left(\frac{1}{20}\right)+75\left(\frac{1}{20}\right)\\ +80\left(\frac{1}{20}\right)+85\left(\frac{1}{20}\right)+90\left(\frac{1}{20}\right)+95\left(\frac{1}{20}\right)+100\left(\frac{1}{20}\right)\end{array}\right\} \\ =52.5 \end{gathered}$$

Thus, the mean of x is 52.5

On an average total score for a single contestant playing is 52

c.

The formula for the variance of x is given below:

${\mathit{蟽}}^{\mathbf{2}}\mathbf{=}\mathit{E}{\left(x-渭\right)}^{\mathbf{2}\mathbf{}}\mathit{P}\left(x\right)$

Therefore,

$\left\{\begin{array}{c}{\left(5-52.5\right)}^2\left(0.5\right)+{\left(10-52.5\right)}^2\left(0.5\right)+{\left(15-52.5\right)}^2\left(0.5\right)\\ +{\left(20-52.5\right)}^2\left(0.5\right)+{\left(25-52.5\right)}^2\left(0.5\right)+{\left(30-52.5\right)}^2\left(0.5\right)\\ +{\left(35-52.5\right)}^2\left(0.5\right)+{\left(40-52.5\right)}^2\left(0.5\right)+{\left(45-52.5\right)}^2\left(0.5\right)\\ +{\left(50-52.5\right)}^2\left(0.5\right)+{\left(55-52.5\right)}^2\left(0.5\right)+{\left(60-52.5\right)}^2\left(0.5\right)\\ +{\left(65-52.5\right)}^2\left(0.5\right)+{\left(70-52.5\right)}^2\left(0.5\right)+{\left(75-52.5\right)}^2\left(0.5\right)\\ \begin{array}{c}+{\left(80-52.5\right)}^2\left(0.5\right)+{\left(85-52.5\right)}^2\left(0.5\right)+{\left(90-52.5\right)}^2\left(0.5\right)\end{array}\\ +{\left(95-52.5\right)}^2\left(0.5\right)+{\left(100-52.5\right)}^2\left(0.5\right)\end{array}\right\}$

The standard deviation is,

$\begin{array}{l}蟽=\sqrt{{蟽}^2}\\ =\sqrt{831.25}\\ =2883\end{array}$

Thus, the standard deviation of x is 28.83.

Here, the wheel is assumed to be 鈥渇air鈥�, that is the outcomes of the wheel are equally likely. Therefore, the distribution is uniform distribution because the continuous random variables appear to have the equally likely outcomes over the range of values. Since, the uniform distribution is not bell or mound shaped. Chebychev鈥檚 rule can be applied to describe the data.

Thus, the Chebychev鈥檚 rule is used in order to find the range of values.

At least$\frac{3}{4}$of the observations are lies between two standard deviations.

At least of the observations are lies between two standard deviations.

The observations will fall within two standard deviations of the mean is,

Thus, at least of the data the range of values for wheel falls in the interval is (-5.16,110.16).

Here, all observations are lie between two standard deviations of the mean. Hence, it is clear that, x is likely to fall in any one interval with the equal length.

d.

If the player spins the wheel twice, the total outcomes are as follows:

Total outcomes=400

The sample space is as follows:

$N\left(S\right)=\left\{\begin{array}{ccccc}(5,5)& \left(10,5\right)& \left(15,5\right)& 鈰�& \left(100,5\right)\\ \left(5,10\right)& \left(10,10\right)& \left(15,10\right)& 鈰�& \left(100,10\right)\\ 鈰�& 鈰�& 鈰�& 鈰�& 鈰�\\ \left(5,100\right)& \left(10,100\right)& \left(15,100\right)& 鈰�& \left(100,100\right)\end{array}\right\}$

Since, there are 400 possible outcomes which are equally likely, the probability of choosing any of the outcome is clearly $\frac{1}{400}=0.025$.

The probability distribution of x is obtained below:

Let x be the sum of the two numbers in each sample.

There is one sample in the sample space with the sum of 10.

role="math" localid="1663303082220" $$\begin{gathered} P\left(x=10\right)=\frac{1}{400} \\ =0.005 \end{gathered}$$

There are two samples in the sample space with the sum of 15.

role="math" localid="1663303088709" $$\begin{gathered} P\left(x=15\right)=\frac{2}{400} \\ =0.005 \end{gathered}$$

There are three samples in the sample space with the sum of 15.

role="math" localid="1663303094451" $$\begin{gathered} P\left(x=15\right)=\frac{2}{400} \\ =0.0075 \end{gathered}$$

Assume the wheel to be fair. If the total of the player鈥檚 spin exceeds 100, then set total score to 0. Then the probability for is obtained by adding all the probabilities and subtract it from the total probability 1.

That is,

$$\begin{gathered} P\left(x=0\right)=1-0.475 \\ =0.5250 \end{gathered}$$

Similarly, for the remaining values is shown in the probability distribution table.

e.

Here, the wheel is assumed to be fair. That is, the outcomes of the wheel are equally likely.