Question

FDA report on pesticides in food. Periodically, the Food and Drug Administration (FDA) produces a report as part of its pesticide monitoring program. The FDA’s report covers a variety of food items, each of which is analysed for potentially harmful chemical compounds. The FDA recently reported that no pesticides were found in 85% of the domestically produced milk, dairy, and egg samples (Pesticide Monitoring Program: The fiscal Year 2012 Pesticide Report, U.S. Food and Drug Administration). Consider a random sample of 800 milk, dairy, and egg items analyzed for the presence of pesticides.

1. Compute, $\mu$and$\sigma$ for the random variable x, the number of dairy-related items showed no trace of pesticide.
2. Based on a sample of 800 dairy-related items, would you likely observe less than half without any traces of pesticide? Explain.

Short answer

1. The value of $\mu$and $\sigma$is 120 and 10.
2. The probability of less than half of without any traces of pesticide is 1.

Step-by-step solution

(a) Given the information

It is given that the sample of size is 800, and the probability of success is 85%.

i.e., $$\begin{gathered} \mathrm{n}=800 \\ \mathrm{p}=85\%=0.85 \end{gathered}$$

The probability density function of the binomial distribution is

$p\left(X=x\right){=}^n{C}_r{p}^x{q}^{n-x}$

The mean of the binomial distribution is

$$\begin{gathered} \mu =np \\ =800\times 0.15 \\ =120 \end{gathered}$$

The variance of the binomial distribution is

$$\begin{gathered} {\sigma }^2=npq \\ =800\times 0.15\times 0.85 \\ =102 \end{gathered}$$

The standard deviation of the binomial distribution is

$$\begin{gathered} {\sigma }^2=102 \\ =10.1 \end{gathered}$$

(b) To find the probability of less than half without any traces of pesticides

Using normal approximation to the binomial distribution.

$$\begin{gathered} x=\frac{n}{2} \\ =\frac{800}{2} \\ =400 \end{gathered}$$

The z – score can be calculated as

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{400-120}{10} \\ =28 \end{gathered}$$

The probability of z is

$\mathrm{p}\left(\mathrm{z}<28\right)=1....\left(\mathrm{from}\mathrm{standard}\mathrm{table}\right)$