Chapter 6, Problem 17CRP

(a)

To determine

Find the probability that 300 or more would be recycled.

Answer to Problem 17CRP

The probability that 300 or more would be recycled is 0.0166.

Explanation of Solution

Calculation:

Conditions for normal approximation to the binomial:

For a binomial experiment with n number of trails, r number of success, probability of success for each trail p and probability of failure $q=1-p$, the r has binomial distribution which is approximated to a normal distribution if,

• $np>5$
• $nq>5$

Mean:

The mean formula for the binomial distribution using normal approximation is,

$\mu =np$

In the formula n denotes number of trails, and p denotes probability of success.

Standard deviation:

The standard deviation formula for the binomial distribution using normal approximation is,

$\sigma =\sqrt{npq}$

In the formula $q=1-p$, n denotes number of trails, and p denotes probability of success.

Conditions for Continuity correction:

The continuity correction is used for converting the discrete random variable r denoting the number of success to continuous normal random variable x,

• The value x is obtained by subtracting 0.5 from r when r is the left point for an interval. That is, $x=r-0.5$.
• The value x is obtained by adding 0.5 from r when r is the right point for an interval. That is, $x=r+0.5$.

Z score:

The number of standard deviations the original measurement x is from the value of mean $\mu$ is measured using the z-score or z value. The formula for z score is,

$z=\frac{x-\mu }{\sigma }$

In the formula, x is the raw score, $\mu$ is the mean and $\sigma$ is the standard deviation.

Let r denotes the number of cans sold in the area were recycled.

The number of trails is $n=400$, and the probability of success for each trail is $p=0.70$.

Checking conditions:

$\begin{array}{c}np=400\left(0.70\right)\\ =280\\ >5\end{array}$

$\begin{array}{c}nq=n\left(1-p\right)\\ =400\left(1-0.70\right)\\ =400\left(0.30\right)\\ =120\\ >5\end{array}$

It can be observed that two of the conditions $np>5$, $nq>5$ are satisfied by the binomial experiment. It is appropriate to use normal approximation to the binomial.

The mean is,

$\begin{array}{c}\mu =np\\ =400\left(0.70\right)\\ =280\end{array}$

The standard deviation is,

$\begin{array}{c}\sigma =\sqrt{npq}\\ =\sqrt{400\left(0.70\right)\left(1-0.70\right)}\\ =\sqrt{84}\\ =9.1652\end{array}$

The probability that 300 or more would be recycled is,

$\begin{array}{c}P\left(r\ge 300\right)\approx P\left(x\ge 300-0.5\right)\\ =P\left(\frac{x-280}{9.1652}\ge \frac{299.5-280}{9.1652}\right)\\ =P\left(z\ge 2.13\right)\\ =1-P\left(z\le 2.13\right)\end{array}$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 2.13.

• Locate the value 2.1 in column z.
• Locate the value 0.03 in top row.
• The intersecting value of row and column is 0.9834.

The probability is,

$\begin{array}{c}P\left(r\ge 300\right)=1-P\left(z\le 2.13\right)\\ =1-0.9834\\ =0.0166\end{array}$

Hence, the probability that 300 or more would be recycled is 0.0166.

(b)

To determine

Find the probability that between 260 and 300 would be recycled.

Answer to Problem 17CRP

The probability that between 260 and 300 would be recycled is 0.9750.

Explanation of Solution

Calculation:

The probability that between 260 and 300 would be recycled is,

$\begin{array}{c}P\left(260\le r\le 3000\right)\approx P\left(260-0.5\le x\le 300+0.5\right)\\ =P\left(\frac{259.5-280}{9.1652}\le \frac{x-7.5}{2.6972}\le \frac{300.5-280}{9.1652}\right)\\ =P\left(-2.24\le z\le 2.24\right)\\ =P\left(z\le 2.24\right)-P\left(z\le -2.24\right)\end{array}$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.24.

• Locate the value –2.2 in column z.
• Locate the value 0.04 in top row.
• The intersecting value of row and column is 0.0125.

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 2.24.

• Locate the value 2.2 in column z.
• Locate the value 0.04 in top row.
• The intersecting value of row and column is 0.9875.

The probability is,

$\begin{array}{c}P\left(260\le r\le 3000\right)=P\left(z\le 2.24\right)-P\left(z\le -2.24\right)\\ =0.9875-0.0125\\ =0.9750\end{array}$

Hence, the probability that between 260 and 300 would be recycled is 0.9750.