The probability that a normally distributed data with a mean of μ and a standard deviation of σ is less than a value x is given by:
[tex]P(X\ \textless \ x)=P\left(z\ \textless \ \frac{x-\mu}{\sigma} \right)[/tex]
Given that μ = 16.5 and σ = 0.804984 and that the probability that the mean oil-change time being at or below the
sample mean for which there is an area of 0.10 to the left under the
normal curve, then:
[tex]P\left(z\ \textless \ \frac{x-16.5}{0.804984} \right)=0.1\\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-16.5}{0.804984} \right)=P(z\ \textless \ -1.281) \\ \\ \Rightarrow\frac{x-16.5}{0.804984}=-1.281 \\ \\ \Rightarrow x-16.5=-1.281(0.804984)=-1.031 \\ \\ x=-1.031+16.5=15.47[/tex]
