Finding the number of days that should be written on carton milk Carton of milk  can be reserved fresh for $20$ days in average, $\frac13$ from the milk cartons can reserved fresh for $22$ days or more.
standard deviation $\sigma=4.651$

Let's assume that the period of fresh is normaly distribued
What is the number of days that the company should  write on the cartons such that $95\%$ of the cartons will be reserved


My attempt:
$$X\sim N(\mu,\sigma^2)$$
$$E[X]=\mu=20$$
$$\text{Var}[X]=\sigma^2=21.631$$
$$X\sim N(20,21.631)$$
The cumulative distribution function:
$$\Phi\bigg(\frac{x-\mu}{\sigma}\bigg)=\Phi\underbrace{\bigg(\frac{x-20}{4.651}\bigg)}_{\stackrel{table}=1.645}\geq95\%$$
$$\frac{x-20}{4.651}=1.645$$
$$x=7.651+20=27.651$$
$\Longrightarrow$ $28$ days

Is it correct so far?

 A: (1) I'm not sure whether you were expected to find $\sigma$ from the
information that $\mu = 20$ and $P(X > 22) = 1/3.$ 
Assuming a normal distribution, that can be done
as follows:
$$0.33 = P(X > 22) = P(Z > (22 - 20)/\sigma),$$
for $Z \sim Norm(0, 1).$
This implies $2/\sigma = .4399$ or $\sigma \approx 4.643.$
But using tables, there are various ways to round, and this
is very close to 4.651, which I use below.
You want your 'sell-by' date to be when at most 5% of the
milk has spoiled. Thus
$$0.05 = P(X < d) = P(Z < (d - 20)/4.651),$$
which implies $(d - 20)/4.651 = -1.645$ or (next lower
integer) $d = 12$ days.
As you begin to learn how to solve problems with the normal distribution, I think it is important to start with sketches
of the standard normal distribution and of the normal distribution
for the problem at hand. Then show the areas corresponding to
the desired probabilities. Such plots are shown below. 
In each plot, the area to the right of the the green line is .95 and the
area to the right of the orange line is 1/3. Roughly speaking, the time period
represented on the plot at the right is the period during which the milk spoils. You want set the sell-by date before a large
amount of spoilage occurs. 

A: Perfectly right.
You have a normal distibution with mean equals to 20 and standard deviation 4.651.
$$X\sim N(20,4.651)$$
Next step is to calculate the $95^{th}$ percentile of aforementioned normal distribution as follow:
$$P_{95^{th}} = 20+1.644853*4.651=27.6502$$
Where $1.644853$ is the $95^{th}$ percentile of standard normal distribution. I have assumed that you already know how to calculate the percentiles for a standard normal distribution.
