Chapter 6, Problem 11PT

a.

To determine

To find: The probability that at least 10 eggs are unbroken.

Answer to Problem 11PT

The probability is 0.96.

Explanation of Solution

Given:

The probability distribution is:

value	0	1	2	3	4
Probability	0.78	0.11	0.07	0.03	0.01

Calculation:

If at least 10 eggs are unbroken, it means at most 2 are broken.

The probability can be calculated as:

  $\begin{array}{c}P\left(Y=2\right)=P\left(Y=0\right)+P\left(Y=1\right)+P\left(Y=2\right)\\ =0.78+0.11+0.07\\ =0.96\end{array}$

Thus, the required probability is 0.96

b.

To determine

To find: The mean and interpret it

Answer to Problem 11PT

The mean is 0.38

Explanation of Solution

Calculation:

The mean can be calculated as:

  $\begin{array}{c}\text{Mean =}{\displaystyle \sum y\times P\left(y\right)}\\ =0\left(0.78\right)+1\left(0.11\right)+2\left(0.07\right)+3\left(0.03\right)+4\left(0.01\right)\\ =0.38\end{array}$

The expected number of broken eggs are 0.38.

c.

To determine

To find: The standard deviation and interpret it

Answer to Problem 11PT

The standard deviation is 0.822.

Explanation of Solution

Calculation:

The standard deviation is calculated as follows:

  $\begin{array}{c}\sigma \text{=}\sqrt{{\displaystyle \sum y^2\times P\left(y\right)}-{\left({\displaystyle \sum y\times P\left(y\right)}\right)}^2}\\ =\sqrt{0^2\times 0.78+1^2\times 0.11+\mathrm{....}+4^2\times 0.01-{\left(0.38\right)}^2}\\ =0.822\end{array}$

The number of broken eggs are expected to differ by 0.822 from the mean of 0.38 eggs.

d.

To determine

To find: The probability that inspector find out the at least 2 broken eggs in one of the three cartons.

Answer to Problem 11PT

The probability is 0.295.

Explanation of Solution

Calculation:

The probability of getting at least 2 broken eggs is:

  $\begin{array}{c}P\left(Y\ge 2\right)=P\left(Y=2\right)+P\left(Y=3\right)+P\left(Y=4\right)\\ =0.07+0.03+0.01\\ =0.11\end{array}$

Now,

  $\begin{array}{l}P\left(X=1\right)=0.11{\left(1-0.11\right)}^{1-1}=0.11\\ P\left(X=2\right)=0.11{\left(1-0.11\right)}^{2-1}=0.0979\\ P\left(X=3\right)=0.11{\left(1-0.11\right)}^{3-1}=0.087131\end{array}$

Thus, the required probability is:

  $\begin{array}{c}P\left(1\le X\le 3\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)\\ =0.11+0.0979+0.087\\ =0.295\end{array}$