91Ó°ÊÓ

Let variable $\mathrm{X}$has normal distribution with mean $\mathrm{μ}$and variance ${\mathrm{σ}}^2$. Then probability density operation is determined by:

$f\left(x\right)=\frac{1}{\sqrt{2\mathrm{π}}\mathrm{σ}}{e}^{-\frac{1}{2}\frac{(x-\mathrm{μ}{)}^2}{{\mathrm{σ}}^2}},x∈R.$

Let $X$ is a variable that describes the measurement of time a CD player lasts. Since the life of CD player is normally distributed with mean $\mathrm{μ}=4.1$ and standard deviation of $\mathrm{σ}=1.3$ years, then our variable $\mathrm{X}$ also has a normal distribution, $X~N(4.1,1.3)$.

We need to define the probability function for $X$and sketch the area of probability that $X$is between $2.8$and $6$years.

The graph is given bellow and the shaded area is probability $P(2.8<X<6)$.

We need to find the probability that CD player will last between $2.8$and $6$years, e.c.$P(2.8<X<6)$.

We have following:

$P(2.8<X<6)={∫}_{2.8}^{6}f\left(x\right)dx$

Substitute the given expression,

$={∫}_{2.8}^6\frac{1}{\sqrt{2\mathrm{π}}1.3}{e}^{-\frac{1}{2}\frac{(x-4.1{)}^2}{1.3^2}}dx$

$={∫}_{2.8}^6\frac{1}{\sqrt{2\mathrm{π}}1.3}{e}^{-\frac{1}{2}{\left(\frac{x-4.1}{1.3}\right)}^2}dx$

$\left|u=\frac{x-4.1}{1.3},du=\frac{1}{1.3}dx\right|$

$={∫}_{-1}^{1.46}\frac{1}{\sqrt{2\mathrm{π}}}{e}^{-\frac{u^2}{2}}du$

$=NORMSDIST(1.46)-NORMSDIST(-1)$

$=0.93-0.16$

We get,

$=0.77\text{.}$