Normal Distribution mean and standard deviation unknown The problem is this:
The daily rainfall in a city is normally distributed with mean μ mm and standard deviation σ mm. The rainfall on one day is independent of the rainfall on any other day. On a randomly selected day, there is a 5% chance that the rainfall is more than 10.2 mm. In a randomly selected seven-day week, there is a probability of 0.025 that the mean daily rainfall is less than 6.1 mm. Find the values of μ and σ.
I have been able to develop expressions for the Z-scores easily enough but have difficulty with the next steps in solving for $\mu$ and $\sigma$.
The textbook gives the answers as $\mu$=7.37 and $\sigma$=1.72
 A: @AlexanderGeldhof has given useful clues toward the solution in Comments. This is a
slightly more detailed presentation.
First, you know that rainfall on a randomly chosen day is 
$X \sim \mathsf{Norm}(\mu, \sigma).$ 
Second, you know that the average rainfall during a randomly
chosen week is $\bar X \sim \mathsf{Norm}(\mu, \sigma/\sqrt{7}).$
Then you are given
$$0.05 = P(X > 10.2) = P\left(Z = \frac{X-\mu}{\sigma} > \frac{10.2-\mu}{\sigma}\right),$$
where $Z$ is standard normal. Thus $\frac{10.2-\mu}{\sigma} = 1.645,$ from
the following computation in R, where qnorm is the quantile function (inverse CDF) of a standard normal distribution. You can also find the value 1.645 using printed normal CDF tables.
qnorm(.95)
[1] 1.644854

Moreover, you are given
$$.025 = P(\bar X < 6.1) = P\left(Z = \frac{\bar X - \mu}{\sigma/\sqrt{7}} < \frac{6.1 - \mu}{\sigma/\sqrt{7}}\right),$$
so that $\frac{6.1 - \mu}{\sigma/\sqrt{7}} = -1.96.$
qnorm(.025)
[1] -1.959964

Now you have two equations in the two unknowns $\mu$ and $\sigma,$ which you
can solve to get numerical values. Except (possibly) for minor rounding discrepancies, your solutions should match the answers provided.
Note: In general, if any two quantiles of a normal distribution are
provided, then they may be used to find its mean $\mu$ and standard deviation $\sigma.$
