91Ó°ÊÓ

Given data:

$\mathrm{μ}=8$,

$m=0.125$

The exponential distribution is defined as follows:

$\mathrm{P}(\mathrm{X}<\mathrm{x})=1-{\mathrm{e}}^{-\mathrm{mx}}$

The required probability can be estimated using the formula,

$\mathrm{P}(\mathrm{x}<1)=1-{\mathrm{e}}^{-0.125×1}$

$\mathrm{P}(\mathrm{x}<1)=1-0.8825$

$\mathrm{P}(\mathrm{x}<1)=0.1175$

$\mathrm{P}(\mathrm{x}<1)=0.1175$is the probability that a light bulb will last less than one year.

Given data:

$\mathrm{μ}=8$

$m=0.125$

The exponential distribution is defined as follows:

$\mathrm{P}(\mathrm{X}<\mathrm{x})=1-{\mathrm{e}}^{-\mathrm{mx}}$

The required probability can be estimated using the formula:

$\mathrm{P}(6<\mathrm{x}<10)=\mathrm{P}(\mathrm{x}<10)-\mathrm{P}(\mathrm{x}<6)$

$\mathrm{P}(6<\mathrm{x}<10)=1-{\mathrm{e}}^{-0.125×10}-\left(1-{\mathrm{e}}^{-0.125×6}\right)$$\mathrm{P}(6<\mathrm{x}<10)=0.4784-0.2865$

$\mathrm{P}(6<\mathrm{x}<10)=0.1859$

Given data:

$\mathrm{μ}=8$,

$m=0.125$

The exponential distribution is defined as follows:

$\mathrm{P}(X<x)=1-e^{-mx}$

$\mathrm{P}(x>k)=1-\left(1-e^{-0.125×k}\right)$

localid="1653537354167" $0.70=e^{-0.125×k}$

Taking log on both sides,

$In(0.70)=In\left({\mathrm{e}}^{-0.125×\mathrm{k}}\right)$

$-0.125\mathrm{k}=-0.3567$

$\mathrm{k}=2.85$

Given data:

$\mathrm{μ}=8$

$m=0.125$

The given probability is $\mathrm{P}(x<k)$

$\mathrm{P}(\mathrm{x}<\mathrm{k})=1-{\mathrm{e}}^{-0.125\mathrm{k}}$

$0.02=1-{\mathrm{e}}^{-0.125\mathrm{k}}$

${\mathrm{e}}^{-0.125\mathrm{k}}=0.98$

Taking log on both sides,

${\mathrm{e}}^{-0.125\mathrm{k}}=0.98$

$-0.125\mathrm{k}=-0.0202$

$\mathrm{k}=0.1616$

The given probability is

$P(x<8âˆ\pounds x>7)=1-P(x<8âˆ\pounds x>7)$.

The memory less exponential distribution can be calculated as follows:

$\mathrm{P}(\mathrm{x}<\mathrm{t}+\mathrm{r}âˆ\pounds \mathrm{x}>\mathrm{t})=\mathrm{P}(\mathrm{x}>\mathrm{r})$$\mathrm{P}(\mathrm{x}<8âˆ\pounds \mathrm{x}>7)=1-\mathrm{P}(\mathrm{x}>7+1âˆ\pounds \mathrm{x}>7)$$\mathrm{P}(\mathrm{x}<8âˆ\pounds \mathrm{x}>7)=1-\mathrm{P}(\mathrm{x}>1)$$\mathrm{P}(\mathrm{x}<8âˆ\pounds \mathrm{x}>7)=1-[1-\mathrm{P}(\mathrm{x}>1\left)\right]$

$\mathrm{P}(\mathrm{x}<8âˆ\pounds \mathrm{x}>7)=\mathrm{P}(\mathrm{x}<1)$

$\mathrm{P}(\mathrm{x}<8âˆ\pounds \mathrm{x}>7)=1-{\mathrm{e}}^{0.125×1}$

$\mathrm{P}(\mathrm{x}<8âˆ\pounds \mathrm{x}>7)=1-0.8825$

$\mathrm{P}(\mathrm{x}<8âˆ\pounds \mathrm{x}>7)=0.1175$