Chapter 6, Problem 1CRE

a.

To determine

To find: $P\left(X=5\right)$ The probability that at least 10 eggs are unbroken.

Answer to Problem 1CRE

The probability is 0.1.

Explanation of Solution

Given:

The probability distribution is:

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

Formula used:

The formulas to compute the mean and standard deviation are:

  $\begin{array}{l}\text{Mean =}{\displaystyle \sum x\times P\left(x\right)}\\ \sigma \text{=}\sqrt{{\displaystyle \sum x^2\times P\left(x\right)}-{\left({\displaystyle \sum x\times P\left(x\right)}\right)}^2}\end{array}$

Calculation:

Let the missing value be x .

The missing value can be calculated as:

  $\begin{array}{c}0.1+0.2+0.3+0.3+x=1\\ x+0.9=1\\ x=0.1\end{array}$

  $P\left(X=5\right)$ can be calculated as:

  $\begin{array}{c}P\left(X=5\right)=x\\ =0.1\end{array}$

The required probability is 0.1

b.

To determine

To find: The probability that both of the patients scores either 1 or 2

Answer to Problem 1CRE

The probability is 0.09.

Explanation of Solution

Calculation:

The probability that both of the patients scores either 1 or 2 is computed as:

  $\begin{array}{c}P\left(1\text{ or 2}\right)=P\left(X=1\right)+P\left(X=2\right)\\ =0.1+0.2\\ =0.3\end{array}$

The probability that both of the patients scores either 1 or 2 is computed as:

  $\begin{array}{c}P\left(1\text{ or 2}\right)=P\left(1\text{ or 2}\right)\times P\left(1\text{ or 2}\right)\\ =0.3\times 0.3\\ =0.09\end{array}$

Thus, the probability is 0.09.

c.

To determine

To find: The mean and standard deviation.

Answer to Problem 1CRE

The mean and standard deviation are 3.1 and 1.14 respectively.

Explanation of Solution

Calculation:

The mean can be calculated as:

  $\begin{array}{c}\text{Mean =}{\displaystyle \sum x\times P\left(x\right)}\\ =1\left(0.1\right)+2\left(0.2\right)+3\left(0.3\right)+4\left(0.3\right)+5\left(0.1\right)\\ =3.1\end{array}$

The standard deviation is calculated as follows:

  $\begin{array}{c}\sigma \text{=}\sqrt{{\displaystyle \sum x^2\times P\left(x\right)}-{\left({\displaystyle \sum x\times P\left(x\right)}\right)}^2}\\ =\sqrt{1^2\times 0.1+2^2\times 0.2+\mathrm{....}+5^2\times 0.2-{\left(3.1\right)}^2}\\ =1.1358\end{array}$

The required mean and standard deviation are 3.1 and 1.14 respectively.