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.

A similar problem is given at brainly.com/question/24261244

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2 years ago