Mathematics: he FDAL level for insect filth in chocolate is 0.6 insect fragments (larvae, eggs, body parts, and so on) per grams. Suppose that there is an average of 0.57 insect fragments per gram of chocolate bar and the distribution is following a Poisson distribution. Compute the probabilities that the number of insect fragments in a 50-gram sample of chocolate bar are:

Well, 6 x 90 = 540So to answer your question, there is only one zero....

he FDAL level for insect filth in chocolate is

Ответ:

madisyn13

12.04.2021

Step-by-step explanation:

$\text{To find :} \\ \\ 1.) \ \ \text{More than 20} \\ \\ 2) \text{Less than 30 } \\ \\ 3) \ \text{Within the range of }( \mu - \sigma , \mu +\sigma)$

$\ \ \ Since \ \ X \ \text{follows a poisson distribution and;} \\ \\ X\sim Poisson (0.57) per \ gram \\ \\ X \sim Poisson (28.5) \ per \ 50 \ grams \\ \\$

$1) \ \ \ \text{The probability } P(X20) = \sum \limits ^{50}_{x=21} \dfrac{e^{-28.5}\times 28.5^x}{x!} \\ \\ \mathbf{ = 0.9389}$

2) P( X

$\text{3) Since then mean}\ \mu = 28.5 \\ \\ \sigma =\sqrt{28.5} = 5.339 \\ \\ \mu - \sigma = 28.5 -5.339= 23.161 \\ \\ \mu + \sigma = 28.5 + 5.339 = 33.839$

∴

$P(23.161 < X< 33.839) = \sum \limits ^{33}_{24} \dfrac{e^{-28.5}\times 28.5^x}{x! } \\ \\ = P(X\le 33) - P( X \le 23) \\ \\ = 0.82678 - 0.1750 \\ \\ \mathbf{= 0.6516}$

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