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There is a​ 10% chance of 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 with mu subscript x overbar baseline equals 16.5μx=16.5 and sigma subscript x overbar baseline equals 0.804984 .σx=0.804984. use technology to find this value for x overbarx. x overbarxequals=15.515.5

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nobillionaireNobley
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]

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