91影视

The given table show the probability distribution for Y .

Consider the missing probability of $X.$

Consider the following for a valid probability model:

The probability of each outcome should be one.

All probability should be between 0 and 1. (including both).

Now,

From the dining table,

The total of all probabilities must equal one.

$0.1+0.2+0.3+0.3+x=1$

Combine phrases that are similar:

$0.9+x=1$

Subtract $0.9$from both sides of the equation:

$x=0.1$

We know that

The missing probability denotes the possibility that the random variable Y equals 5.

$P(Y=5)=0.1=10\%$

Thus,

There is a $10\%$probability that a randomly picked patient's pain score will be equal to $5.$

The given table show the probability distribution for Y .

From above result

Now,

Note that

The likelihood of the pain score being equal to one,

$P(Y=1)=0.1$

The likelihood of the pain score being equal to two,

$P(Y=2)=0.2$

We know that a single patient's pain score cannot be equal to two separate numbers.

As a result, the two occasions are mutually exclusive.

For mutually exclusive events, use the addition rule:

$\begin{array}{rcl}P(Y& 鈮�& 2)=P(Y=1)+P(Y=2)\\ & =& 0.1+0.2\\ & =& 0.3\\ & =& 30\%\end{array}$

Thus,

There are $30\%$Chances that the patient chosen at random will have a pain score of at least 2.

The given table show the probability distribution for Y .

From above part (a) result

The sum of each possibility x multiplied by its probability P(x) is the expected value (or mean). .

$\begin{array}{rcl}渭& =& 鈭�xP\left(x\right)\\ & =& 1脳P(Y=1)+2脳P(Y=2)+3脳P(Y=3)+4脳P(Y=4)+5脳P(Y=5)\\ & =& 1脳0.1+2脳0.2+3脳0.3+4脳0.3+5脳0.1\\ & =& 3.1\end{array}$

Now,

The expected value of the squared departure from the mean is called variance.

$\begin{array}{rcl}{蟽}^2& =& 鈭�(x-渭{)}^{2}P(x)\\ & =& (1-3.1{)}^2脳P(Y=1)+(2-3.1{)}^2脳P(Y=2)+(3-3.1{)}^2脳P(Y=3)+(4-3.1{)}^2脳P(Y=4)+(5-3.1{)}^2脳P(Y=5)\\ & =& (1-3.1{)}^2脳0.1+(2-3.1{)}^2脳0.2+(3-3.1{)}^2脳0.3+(4-3.1{)}^2脳0.3+(5-3.1{)}^2脳0.1\\ & =& 1.29\end{array}$

We are aware of this.

The square root of the variance is the standard deviation.

Thus,

$\begin{array}{rcl}蟽& =& \sqrt{{蟽}^2}\\ & =& \sqrt{1.29}\\ & 鈮�& 1.1358\end{array}$