The noise level of a washing machine is a aleator variable with the expected value $44$ dB and the standard deviation $5$ dB. Admitting that aproximation is normal, find the probability that the expected value of the noise is higher to $48$ dB, in a sample with waist $10$ washing machines.
I find that the standard deviation $ \sigma = \sqrt{ E[(X-\mu)^{2}]}$, where $\mu = E[X]$ and $ \mu = 44 dB$ and $ \sigma = 5 dB$, but I don’t know how to continue. Thank you!
Use the Central Limit Theorem: $P( \bar x > 48 ) = P\left(z > \dfrac{48 - \mu}{\dfrac{\sigma}{\sqrt{n}}}\right) = P\left(z >\dfrac{48-44}{\dfrac{5}{\sqrt{10}}}\right)$. Can you continue here?