A tool weight is distributed normally with mean = $2265.4$. Given that 14% of the tools' weight are above 2278.36. what is the standard deviation? 
A tool weight is distributed normally with mean = $2265.4$. Given that 14% of the tools' weight are above 2278.36. what is the standard deviation?

Here the solution:

denote $X$ as tool's weight. $X$ ~ $N(2265.4, \sigma^2)$
$P(X > 2278.36) = 0.14.$
$1 - P(X < 2278.36) = 0.14.$
$P(X < 2278.36) = 0.86.$
$(**) \space\space P(Z < \frac{2278.36 - 2265.4}{\sigma} = 1.08) = 0.86$
$\sigma = \frac{2278.36 - 2265.4}{1.08} = 12.$

Can you please expalin me the solution? I can't understand what's going on $(**)$. what is this $1.08$?
Thanks in advance.
 A: On the line you don't understand, you are using the standardization of your random variable $X$ to find your standard deviation value $\sigma$. Recall that in order to standardize a Normally distributed random variable, you subtract its mean, and divide by its standard deviation. Essentially your formula is 
$$
Z = \dfrac{x-\mu}{\sigma}
$$
where $x$ denotes your corresponding value, $\mu$ denotes your mean and $\sigma$ your standard deviation. Then you can look up your Z-score value in the Normal distribution table, and get an approximate tail probability for your random variable. 
In this case, you know that $P(X > 2278.36) = 0.14$. We can "work backwards" essentially, and by taking the complementary $1-P(X<2278.36)$ we then have $P(X<2278.36)=0.86$. Then note that 
$$P(X<2278.36) = P\big(Z = \dfrac{x-\mu}{\sigma}\big) = 0.86$$
Now look in your Normal distribution table for the $Z$ value that corresponds to a probability of $0.86$, it is around $1.08$. 
So then we know $1.08 = \dfrac{x-\mu}{\sigma}$, and as you know $x$ and $\mu$ you simply solve for $\sigma$.
