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:
And standard deviation of:
![\sqrt{V(X)} = \sqrt{np(1-p)}](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D)
Normal probability distribution
Z-score formula is used, which, in a set with mean
and standard deviation
, the z-score of a measure X is given by:
- 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](https://tex.z-dn.net/?f=p%20%3D%200.65)
- 80 samples, thus
![n = 80](https://tex.z-dn.net/?f=n%20%3D%2080)
The mean of X is of:
![E(X) = np = 0.65(80) = 52](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np%20%3D%200.65%2880%29%20%3D%2052)
The standard deviation of X is of:
![\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{80(0.65)(0.35)} = 4.2661](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B80%280.65%29%280.35%29%7D%20%3D%204.2661)
Using the approximation and continuity correction, the probability is
, which is the p-value of Z when X = 39.5. Thus:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{39.5 - 52}{4.2661}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B39.5%20-%2052%7D%7B4.2661%7D)
![Z = -2.93](https://tex.z-dn.net/?f=Z%20%3D%20-2.93)
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