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I am struggling with the following question:

A company which produces $1L$ beverages adjusts their machines in a way that the filling quantity is normally distributed. The mean is $\mu=995\,\text{cm}^3$ and the standard deviation is $\sigma = 5\,\text{cm}^3$.

To prevent manipulations, authorities take samples. If in a sample of five cans at least four cans contain more than $997\,\text{cm}^3$ and the fifth contains more than $995 \,\text{cm}^3$, nothing will be queried. With what probability are those manipulations discovered (1) in one sample (2) in 10 samples.

I tried the following:

$$\mu=995cm^3 \space\space\space\space \sigma=5cm^3$$

$$P(X>997cm^3)=1-\int_{-\infty}^{997}\frac{1}{{\sigma \sqrt {2\pi } }}e^{ - \frac{(x - \mu)^2}{2\sigma ^2}}\,\mathrm dx$$

$$P(X>995cm^3)=1-\int_{-\infty}^{995}\frac{1}{{\sigma \sqrt {2\pi } }}e^{ - \frac{(x - \mu)^2}{2\sigma ^2}}\,\mathrm dx$$

(1) $P=P(X>997cm^3)^4*P(X>995cm^3)$ since four have to contain more than $997 cm^3$ and one has to contain more than $995 cm^3$.

(2) $P=10*(1)$ since we have ten times more tries now.

I get for (1) $0.7%$ and for (2) $7%$ but according to the solutions' book I should get a percentage of about $98$ for (1) and $~100%$ for (2).

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In (1), the probability you are calculating is the probability that nothing is queried. The question is the reverse, what is the probability that manipulations are discovered?

In (2), what you are looking for is the complement to no queries in all samples. Hence, if $p_{(1)}$ is the probability that nothing will be queried in one sample, you want $1-\left[p_{(1)}\right]^{10}$. If you just multiply it by 10 you could very well end up with a 'probability' which is larger than 1.

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