A particular breed of hen produces eggs with a mean mass of $60$ grams and a standard deviation of $4$ grams and mass is found to be normally distributed.
Eggs are classified as small if their mass is less than $55$g.
a. Find the proportion of eggs that are not classified as small.
I've worked out the first part. I did $(55-60)/4=-1.25$. Then I found the $Z$-score which was $0.89435$, and then $\times100$ to get $89.435\%$.
b. Find the range of masses, symmetrical about the mean, such that $50\%$ of the eggs are within this range.
I don’t understand how to get to the second part; I just don’t understand how you find the range of masses that are within $50\%$. I understand that it will be $25\%$ on each side, I just don’t know how to work it out.
So the entire HW problem here is about how to work with the standard normal distribution and it's CDF.
The key to solving these problems is knowing what the probability associated with each $z-$score represents. For example, $z=1.96$ is the $97.5$th percentile, which is used to construct confidence intervals and two-sided rejection regions with confidence/significance level $0.05$.
In general, if you have some interval $(a,b)$ and want to know $P(X \in (a,b))$ then if $X$ is normally distributed, you calculate the $z-$score for $a$ and $b$ so you can look up the correct probabilities:
$z_a = \frac{a-\mu}{\sigma}, z_b = \frac{b-\mu}{\sigma}$
The z-score gives the number of standard deviations away from the mean. The probability that a normally distributed random variable is within $z$ standard deviations of it's mean is: $P(X \leq \mu + z\sigma) - P(X \leq \mu - z\sigma)$