This is one of our homework problems for Statistics. I'm really stuck on this one, and it would be great if someone could show me how to solve this, or at least set up the equation.
The weights of chickens on old McDonald’s farm are normally distributed with a standard deviation of 4.2 lb. Find the mean weight of old McDonald’s chickens if only 4% weigh less than 20 lbs.
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Hint: The equation is $P(X \leq 20)=\Phi\left( \frac{20-\mu}{4.2} \right)=0.04$, where $\Phi(.)$ is the standard normal distribution. Solve the equation for $\mu$. The value for $\Phi^{-1}(0.04)$ can be looked up in the table for the standard normal distribution.
Additional:
It might be helpful, that $1-\Phi(z)=\Phi(-z)$. Insert the value for $\Phi(-z)=0.04$ leads to $1-\Phi(z)=0.04$. This equation can be transformed to $\Phi(z)=0.96$. Find for the probability 0.96 (nearest value) the corresponding z-value. This value has to be multiplied by (-1) and you have $\Phi^{-1}(0.04)$
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You have to use the fact that the weights are normally distributed.
You're given the standard deviation, and a value for the left 4% of the distribution.
You can calculate (or look up) how many standard deviations away from the mean you are at 4%.
Given this, you multiply the number of standard deviations from the mean by the standard deviation (in lbs) to see how many lbs away from the mean you are.
You know one end (20 lbs) and you can now find the other end, which is the mean.
Let the weight of the chickens be $X$. Then $X \sim N(\mu,\sigma^2)$ where $\sigma = 4.2$ is the standard deviation of the weights and $\mu$ is the mean of the weights. Then, the statement you're given is $P(X < 20) = 0.04$.
Now, rewrite $P(X<20) = P(X-\mu < 20-\mu) = P( \frac{X-\mu}{\sigma} < \frac{20-\mu}{\sigma}) = P(Z < \frac{20-\mu}{\sigma})$ where $Z \sim N(0,1)$. Then, look up $c$ in a standard normal table such that $P(Z<c) = 0.04$ and then note $\frac{20-\mu}{\sigma} = c$. Solve for $\mu$, and you are done.