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Here is an exercise, I tried to solve. It is a long since, I study statistics, so I would really appreciate if someone could check the answer/solution I give below.

Exercise: Assume we are given a sample of $160$ cows. The mean value of the milk production is $ \mu= 100 \textrm{kg},$ and the standard deviation is $ \sigma= 30 \textrm{kg}.$ We wish to replace the $20\%$ of the cows with the minimum production of milk. Find the quantity of milk the these cows produce.

Solution: We denote by $X$ the random variable of the amount of milk. By the central limit theorem, we can assume that $ X \sim N(100, 30^2) ,$ and thus

$$ \frac{ X - 100}{30} \sim N(0,1)$$

follows the standard normal distribution.

Put $ c$ for the maximum quantity of milk, produced by cows that belong to the worst $20 \%.$ Then we have:

$$ P( X \leq c ) =0.2= 1-0.8=1- \Phi(0.84)= \Phi(-0.84)$$

$$ \Longrightarrow P(Z \leq \frac{ c-100}{30})= \Phi(-0.84) ,$$

where $ \Phi$ is the probability density function of the standard normal distribution.

Therefore we obtain

$$ \frac{c-100}{30} = -0.84 \Longleftrightarrow c=74,8 .$$

In conclusion, the cows that we should replace, are these that produce milk less than $ 74.8 \textrm{kg}. \quad \square$

Could someone check this attempt of solution ?

Thank you in advance!

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    Looks good: In R statistical software `qnorm(.2, 100, 30)` returns 74.75136. That is, the 20th percentile (quantile .2) of $NORM(\mu=100,\sigma=30)$ is about 74.75kg. Your answer is probably as close as you can get with printed normal tables (without interpolation). – 2017-01-23
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    Dear Bruce, thank you very much for cheking my solution. If you want, you could make your comment into an answer, so I can accept it. – 2017-01-25
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    Perhaps not enough substance to Comment for a formal Answer. Knowing it was useful will be sufficient reward on this one. Thanks for your Comment. – 2017-01-25

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