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.

Step by step solution

01

(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

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

02

(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

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

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