I'm still not sure I follow your formulation, but I'll solve according to my understanding and feel free to correct me.
Current situation: you have some fixed quantity $x$ and you have some "noise" (uncertainty in demand or forecast error) $N$. The overall demand is therefore $$Y=x+N.$$ Now, you would to increase your quantity such that the probability of someone asking for more that you have is less than 3% (i.e. 97% service level). Then, you find $m$ such that
$$P(Y>m)<0.03$$
$$P(x+N>m)<0.03$$
$$1-P(N<m-x)<0.03$$
$$0.97<P(N<m-x)$$
$$P(N<1.89)<P(N<m-x)$$
$$1.89<m-x$$
$$m>x+1.89$$
so you need to add 1.89 to your quantity to be sure that in 97% certainty you'll have enough stock.
Desired situation: you have an additional uniformly distributed quantity $U\sim \mathcal U[0,u]$ of between 0 and $u$ items that's added to your stock. So you have:
$Z=x+N+U$ (unrelated to your $z$ in the question), and you need to figure out what's the value of $m$ such that $$P(Z>m)<0.03,$$ exactly as before, only now $Z$ contains another quantity.
Before you could have easily calculated this since you have the normal distribution tables. But now, $P(Z>m)=P(x+N+U>m)=P(N+U>m-x)=1-P(N+U<m-x)$, and $R\triangleq N+U$ is not normally distributed. Its distribution is given by the following convolution:
$$f_R(r) = (f_N*f_U)(r) = \int_{-\infty}^\infty f_U(t) f_N(r-t) dt$$
$$ = \frac{1}{u} \int_0^u f_N(r-t) dt$$
$$ = \frac{1}{u} \int_{r-u}^{r} f_N(t) dt$$
Now, since $f_N(t)$ is a normal density, we get
$$f_R(r) = \frac{1}{u} [\Phi(r)-\Phi(r-u)]$$
To get back to our problem, we have
$$0.97<P(R<m-x)$$
$$0.97<F_R(m-x)$$
$$F^{-1}_R(0.97)<m-x$$
$$m>x+F^{-1}_R(0.97)$$
where $F^{-1}_R(r)$ is the inverse cdf of $R$. We have the density, $f_R(r)$. We need to numerically integrate it to get the cdf and then find the value in which it is equal to $0.97$. Fortunately, we can use MATLAB (or any other software), and get:
The code is:
resolution = 0.01;
u = 5; val = 0.97;
r = -10:resolution:10;
fr = 1/u * ( normcdf(r) - normcdf(r-u) );
dr = r(2)-r(1);
Fr = cumsum(fr*dr);
[~,idx] = min( (Fr-val).^2 );
Fr_inv_at_val = r(idx);
subplot(211); plot(r,fr,'linewidth',2), title('f_R(r)');
subplot(212);
hold off; plot(r,Fr,'linewidth',2),
hold on; plot(r(idx),val,'go','linewidth',2)
title(sprintf('F_R(r), F_R^{-1}(%1.3f)=%1.3f',val,Fr_inv_at_val));
So, the value we want is $m>5.67$, the value we need to add to the current stock, for an arbitrary choice of $u=5$ and certainty of $0.97$.
If I check it by simulation:
tot_exps=10000;
x = 10;
u = 5;
N = randn(tot_exps,1);
U = rand(tot_exps,1)*u;
Y = x+N+U;
m = 5.67;
res = mean(Y>x+m)
the value of res is the average number of times the "customer" ($Y$) wanted more than I kept ($x+m$), then indeed I get values very close to 3%.
EDIT:
According to your comments, here's a general way to solve it, using Excel.
The main equation here is $$P(N>m+c)<0.03$$
$$1-P(N<m+C)<0.03$$
$$0.97<P(N-C<m)$$
Let $R=N-C$, then
$$0.97<F_R(m)$$
$$F_R^{-1}(0.97)<m$$
so, we need to do the following:
a. get $f_R(r)$ for $R=N+C$ where $N$ is standard normal and $C$ is some uniformly distributed variable. Note that I took $C$ instead of $-C$ but due to the uniform distribution it doesn't matter as long as the boundaries are correct.
Similarly to before, $f_R(r)$ is given by:
$$f_R(r) = \frac{1}{u-v} [\Phi(r-v)-\Phi(r-u)]$$
where $u,v$ are the boundaries of the uniform distribution.
b. Get the cdf via $F_R(r)$ by cumulative summation (numerical integration).
c. find the value of $r$ for which $F_R(r)=0.97$, i.e. find $F^{-1}_R(0.97)$.
Now, I'm pretty rusty in Basic so I wrote up something in Excel, but you might be able to do much better. Nevertheless, the code produces the correct result.
The main idea in this file is to write the values for the cdf in Excel cells, and then use a VBA script to calculate $F_R(r)$ in the desired values and look for the correct $r$ that matches $0.97$.
u = 5
v = 0
desired_val = 0.97
max_cell = 2001
dr = Sheet1.Cells(1, 16) ' dr
For i = 1 To max_cell
r = Sheet1.Cells(i, 15)
s = 1 / (u - v) * (WorksheetFunction.Norm_Dist(r - v, 0, 1, True) - WorksheetFunction.Norm_Dist(r - u, 0, 1, True))
Sheet1.Cells(i + 1, 14) = Sheet1.Cells(i, 14) + s * dr
Next i
' now we have the values of the cdf of the random variable R in the N column
' next, we need to find which item in N column (the cdf) is closest to 0.97
chosen_idx = 1
min_diff = 1000 ' some high value
For i = 1 To max_cell
cur = Abs(Sheet1.Cells(i, 14) - desired_val)
If cur < min_diff Then
chosen_idx = i
min_diff = cur
End If
Next i
' now we have the index of the suitable value for which F_R = 0.97. to invert we just take the value from O in this index
chosen = Sheet1.Cells(chosen_idx, 15)
Sheet1.Cells(8, 8) = chosen
I'm not sure if it's allowed to attach files here, but I attach a link to the xlsx file itself:
https://www.dropbox.com/s/1cvfrkna8vbged3/Book1.xlsm?dl=0
