Chapter 7, Problem 21CR

Recycling: Aluminum Cans One environmental group did a study of recycling habits in a California community. It found that 70% of the aluminum cans sold in the area were recycled.

(a) If 400 cans are sold today, what is the probability that 300 or more will be recycled?

(b) If the 400 cans sold, what is the probability that between 260 and 300 will be recycled?

To determine

To find:

The probabilitythat 300 or more cans will be recycled.

Answer to Problem 21CR

Solution:

The probability that 300 or morecans will be recycled is 0.0166

Explanation of Solution

Calculation:

We have random variable x having a normal distribution with mean $\mu$ and standard deviation $\sigma$. Using $n=400,p=0.70,q=1-p=\mathrm{0.30.}$

Now, the mean is

$\begin{array}{l}\mu =np\\ =400*0.70\\ =280\end{array}$

And standard deviation will be

$\begin{array}{l}\sigma =\sqrt{npq}\\ =\sqrt{400*0.70*0.30}\\ =\sqrt{84}\\ =9.165\end{array}$

Now, we calculate $P(r\ge 300)$ using continuity correction factor for normal distribution.

$\begin{array}{l}P(r\ge 300)=P(x\ge r-0.5)\\ =P(x\ge 300-0.5)\\ =P(x\ge 299.5)\end{array}$

By using formula for normal distribution:-

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ z=\frac{299.5-280}{9.165}\\ z=2.13\end{array}$

By using Table 3 from appendix:

$\begin{array}{l}P(x\ge 299.5)=P(z\ge 2.13)\\ =1-P(z\le 2.13)\\ =1-0.9834\\ =0.0166\end{array}$

The probability that 300 or more cans will be recycled is 0.0166.

To determine

To find:

The probability that between 260 and 300 canswill be recycled.

Answer to Problem 21CR

Solution:

The probabilitythat between 260 and 300cans will be recycled is 0.9750.

Explanation of Solution

Calculation:

We have random variable x having a normal distribution with mean $\mu$ and standard deviation $\sigma$. Using $n=400,p=0.70,q=1-p=\mathrm{0.30.}$

Now, we calculate $(260\le r\le 300)$ using continuity correction factor for normal distribution.

$\begin{array}{l}P(260\le r\le 300)=P(r-0.5\le x\le r+0.5)\\ =P(260-0.5\le x\le 300+0.5)\\ =P(259.5\le r\le 300.5)\end{array}$

Now, mean and standard deviation are as $\mu =280,\sigma =9.165$

Further,

$\begin{array}{l}P(259.5\le x\le 300.5)=P\left(\frac{259.5-280}{9.165}\le \frac{x-\mu }{\sigma }\le \frac{300.5-280}{9.165}\right)\\ =P(-2.24\le z\le 2.24)\\ =P(z\le 2.24)-P(z\le -2.24)\\ =0.9875-0.0125\\ \text{}=0.9750\end{array}$

Thus, the desired probability is 0.9750.