Chapter 6, Problem R6.1RE

(a)

To determine

Probability for the event $Y=5$ .

Answer to Problem R6.1RE

Probability,

  $P\left(Y=5\right)=0.1$

Explanation of Solution

Given information:

Table showing probability distribution for Y :

Calculations:

Let the missing probability be x .

Such that

For a valid probability model:

• For each outcome, sum of probabilities should be equal to 1.
• All probabilities should be between 0 and 1 (including both).

Now,

From the table,

Sum of all probabilities needs to be equal to 1.

  $0.1+0.2+0.3+0.3+x=1$

Combine like terms:

  $0.9+x=1$

Subtract 0.9 from both sides:

  $x=0.1$

We know that

The missing probability represents the event that the random variable Y which is equal to 5.

  $P\left(Y=5\right)=0.1=10\%$

Thus,

There are 10% chances that the pain score of a randomly selected patient is equal to 5.

(b)

To determine

Probability for the randomly selected patient has a pain score of at most 2.

Answer to Problem R6.1RE

Probability for pain score of at most 2,

  $P\left(Y\le 5\right)=0.3$

Explanation of Solution

Given information:

Table showing probability distribution for Y :

Calculations:

From Part (a) result,

Table becomes:

Now,

Note that

Probability for pain score being equal to 1,

  $P\left(Y=1\right)=0.1$

Probability for pain score being equal to 2,

  $P\left(Y=2\right)=0.2$

We know that

It is impossible for one patient to have pain score being equal to different values.

Thus,

The two events are mutually exclusive.

Apply addition rule for mutually exclusive events:

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

Thus,

There are 30% chances for the randomly selected patient to have pain score of at most 2.

(c)

To determine

Expected pain score and the standard deviation of the pain score.

Answer to Problem R6.1RE

Expected pain score,

  $\mu =3.1$

Standard deviation of the pain score,

  $\sigma \approx 1.1358$

Explanation of Solution

Given information:

Table showing probability distribution for Y :

Calculations:

From Part (a) result,

Table becomes:

The expected value (or mean) is the sum of the product of each possibility x with its probability $P\left(x\right)$ .

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

Now,

Variance is the expected value of the squared deviation from the mean.

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

We know that

The standard deviation is the square root of the variance.

Thus,

  $\sigma =\sqrt{{\sigma }^2}=\sqrt{1.29}\approx 1.1358$