Question

The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with a mean of 32 and a standard deviation of 4. Round to the second. a) Find the 50 percentile of tire lifetimes. b) Find the tire life that has 70% above it. c) Find the two tire lives that hold the middle 15% Lower Upper d) The tire company wants to guarantee that its tires will last a certain number of miles. What number of miles (in thousands) should the company guarantee so that only 1% of the tires would violate the guarantee? Points possible: 2 This is attempt 1 of3 License

Answer 1

$\mathbf{Solution}$

$\mathbf{given:}\;\mu=32,\;\sigma=4$

$\mathbf{formula:}\;Z=\frac{X-\mu}{\sigma}$

${\color{Blue} \mathbf{a)}}\;50\;percentile\Rightarrow P(Z<?)=0.50$

From Z-table, Lookup for Z-value corresponding to area 0.50 to the left of the normal curve.

$\Rightarrow P(Z<\mathbf{0})=0.50$

$0=\frac{X-32}{4}$

$\therefore X=\mathbf{{\color{Red} 32}}$

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${\color{Blue} \mathbf{b)}}\;70\%\;above\Rightarrow P(Z>?)=0.70$

From Z-table, Lookup for Z-value corresponding to area 0.70 to the right of the normal curve.

$\Rightarrow P(Z>\mathbf{-0.52})=0.70$

$-0.52=\frac{X-32}{4}$

$X=32+(-0.52*4)$

$X=32-2.08$

$X=\mathbf{{\color{Red} 29.92}}$

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${\color{Blue} \mathbf{c)}}\;middle\;15\%\Rightarrow P(?<Z<?)=0.15$

$\mathbf{for\;two \;tail\; test}\;\frac{\alpha }{2}=\frac{1-0.15}{2}=\mathbf{0.425}$

From Z-table, Lookup for Z-value corresponding to area 0.425 to the left of the normal curve.

$\Rightarrow P(Z<-0.19)=0.425$

From Z-table, Lookup for Z-value corresponding to area 0.425 to the right of the normal curve.

$\Rightarrow P(Z>0.19)=0.425$

$\therefore P(\mathbf{-0.19}<Z<\mathbf{0.19})=0.15$

${\color{Blue} \mathbf{Lower}}$

$-0.19=\frac{X-32}{4}$

$X=32+(-0.19*4)$

$X=32-0.76$

$X=\mathbf{{\color{Red} 31.24}}$

${\color{Blue} \mathbf{Upper}}$

$0.19=\frac{X-32}{4}$

$X=32+(0.19*4)$

$X=32+0.76$

$X=\mathbf{{\color{Red} 32.76}}$

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${\color{Blue} \mathbf{d)}}\;only\;1\%\Rightarrow P(Z<?)=0.01$

From Z-table, Lookup for Z-value corresponding to area 0.01 to the left of the normal curve.

$\Rightarrow P(Z<\mathbf{-2.33})=0.50$

$-2.33=\frac{X-32}{4}$

$\therefore X=\mathbf{{\color{Red} 22.68}}$