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$$\left.P(\text{Vebrauch}\le v)=0,88\quad\right|\text{normalisieren mit }\mu=5,99\;;\;\sigma=0,38$$$$\left.\phi\left(\frac{v-5,99}{0,38}\right)=0,88\quad\right|\phi^{-1}(\cdots)$$$$\left.\frac{v-5,99}{0,38}=1,174987\quad\right|\cdot0,38$$$$\left.v-5,99=1,174987\cdot0,38\quad\right|+5,99$$$$\left.v=5,991,174987\cdot0,38\quad\right|\text{ausrechnen}$$$$v\approx6,436$$