Question

The Food and Drug Administration (FDA) produces a quarterly report called Total Diet Study. The FDA’s report covers more than 200 food items, each of which is analyzed for dangerous chemical compounds. A recent Total Diet Study reported that no pesticides at all were found on 65% of the domestically produced food samples. Consider a random sample of 80 food items analyzed for the presence of pesticides. Let X be the number of food items found without any trance of pesticide. Calculate the mean and standard deviation of X. Based on a sample of 80 food items, what is the probability that less than half without any traces of pesticide?

Answer 1

Using the normal approximation to the binomial, it is found that:

• The mean of X is of 52 samples.
• The standard deviation of X is of 4.2661 samples.
• 0.0017 = 0.17% probability that less than half without any traces of pesticide.

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Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

Can be approximated to a normal distribution, with mean given by:  

$E(X) = np$

And standard deviation of:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution  

Z-score formula is used, which, in a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• Each z-score has an associated p-value, which measures the percentile of measure X.

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• 65% of samples had no pesticide, thus $p = 0.65$
• 80 samples, thus $n = 80$

The mean of X is of:

$E(X) = np = 0.65(80) = 52$

The standard deviation of X is of:

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{80(0.65)(0.35)} = 4.2661$

Using the approximation and continuity correction, the probability is $P(X < 40 - 0.5) = P(X < 39.5)$, which is the p-value of Z when X = 39.5. Thus:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{39.5 - 52}{4.2661}$

$Z = -2.93$

$Z = -2.93$ has a p-value of 0.0017, thus:

0.0017 = 0.17% probability that less than half without any traces of pesticide.

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