Question

Knees Patients receiving artificial knees often experience pain after surgery. The pain is measured on a subjective scale with possible values of 1 (low) to 5 (high). Let Y be the pain score for a randomly selected patient. The following table gives the probability distribution for Y.

Value	1	2	3	4	5
Probability	0.1	0.2	0.3	0.3	??

a. Find $P(Y=5)$. Interpret this value.

b. Find the probability that a randomly selected patient has a pain score of at most 2 .

c. Calculate the expected pain score and the standard deviation of the pain score.

Short answer

1. The required probability of the event is $P(Y=5)=0.1$.
2. There is a $30\%$chance that the randomly selected patient will have a pain score of at least $2$.
3. The required expected pain score \mu=3.1 and Standard deviation of the pain score is
   \sigma \approx 1.1358
   
   

Step-by-step solution

Part (a) Step 1: Given information

The given table show the probability distribution for Y .

Value	1	2	3	4	5
Probability	0.1	0.2	0.3	0.3	??

Part (a) Step 2: Calculation

Consider the missing probability of $X.$

Value	1	2	3	4	5
Probability	0.1	0.2	0.3	0.3	??

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 $5.$

Part (b) Step 1: Given information

The given table show the probability distribution for Y .

Value	1	2	3	4	5
Probability	0.1	0.2	0.3	0.3	??

Part (b) Step 2: The likelihood that a randomly chosen patient has a pain score of no more than 2

From above result

Value	1	2	3	4	5
Probability	0.1	0.2	0.3	0.3	0.1

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:

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

Thus,

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

Part (c) Step 1: Given information

The given table show the probability distribution for Y .

Value	1	2	3	4	5
Probability	0.1	0.2	0.3	0.3	??

Part (c) Step 2: Calculation

From above part (a) result

Value	1	2	3	4	5
Probability	0.1	0.2	0.3	0.3	0.1

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

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

Now,

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

$\begin{array}{rcl}{\sigma }^2& =& \sum (x-\mu {)}^{2}P(x)\\ & =& (1-3.1{)}^2\times P(Y=1)+(2-3.1{)}^2\times P(Y=2)+(3-3.1{)}^2\times P(Y=3)+(4-3.1{)}^2\times P(Y=4)+(5-3.1{)}^2\times P(Y=5)\\ & =& (1-3.1{)}^2\times 0.1+(2-3.1{)}^2\times 0.2+(3-3.1{)}^2\times 0.3+(4-3.1{)}^2\times 0.3+(5-3.1{)}^2\times 0.1\\ & =& 1.29\end{array}$

We are aware of this.

The square root of the variance is the standard deviation.

Thus,

$\begin{array}{rcl}\sigma & =& \sqrt{{\sigma }^2}\\ & =& \sqrt{1.29}\\ & \approx & 1.1358\end{array}$