you are looking for p(45) which is the consumption level separating the bottom 45% from the top 55%.
you are looking for the z-score that has 45% of the area under the normal distribution curve to the left of it.
formula for z-score is z = (x-m)/s
z is the z-score.
x is the raw score.
m is the mean.
s is the standard deviation.
best to use a z-score calculator rather than looking this up in the z-score table.
one calculator that can be used is the one at https://homepage.stat.uiowa.edu/~mbognar/applets/normal.html
using this calculator, you would leave the mean at 0 and the standard deviation at 1 and select p(X < x) and then enter .45 in the red box.
once you hit the return, it will tell you that x = -0.12566 in the white box.
if you left mean at 0 and standard deviation at 1, then that's your z-score.
here's a display of the results from using this calculator.
you would then find the raw score by using the z-score formula of z = (x-m)/s
z is the z-score.
x is the raw score.
m is the raw score mean.
s is the raw score standard deviation.
solve for x to get x = z * s + m.
when z = -.12566 and s = 218 and m = 1050, you would get x = -.12566 * 218 + 1050 = 1022.60612.
that tells you that 1022.60612 kwh separates the bottom 45% from the top 55%.
visually, this looks like what's shown in the following two graphs.
first graph shows 45% in bottom half (area to the left of 1022.60612 kwh)
second graph shows 55% in the top half (area to the right of 2022.60612 kwh).
there's some internal truncating going on in the graphing software, so the raw score shows as 1022.606 rather than 1022.60612 which has a small impact on the displayed results using the graphing calculator.
your solution is that the consumption level separating the bottom 45% from the top 55% is equal to 1022.60612 when using the referenced normal distribution calculator at https://homepage.stat.uiowa.edu/~mbognar/applets/normal.html