Normal distribution - class-XI

In a normal distribution , mean, median and mode are equal. The curve is bell type curve which is symmetric about 

Let $X$ denote the marks secured.

Let $x$ be the annual salary of employees in a large company.

$x$ has $\mu=50000,\sigma=20000$.

We know that for given $x,z=\dfrac{x-\mu}{\sigma}$

We have to find the percent of people earning between $45,000$ and $65,000$

First let us find $P(45000<x<65000)$

For $x=45000,z=\dfrac{45000-50000}{20000}=-0.25$
and for $x=65000,z=\dfrac{65000-50000}{20000}=0.75$

$\therefore P(45000<x<65000)=P(-0.25<z<0.75)$

                                               $=P(z<0.75)-P(z<-0.25)$

                                               $=0.7734-(1-0.5986)$ (from normal distribution table)

                                               $=0.372$

$\therefore P(45000<x<65000)=0.372=37.2%$

Hence the percent of people earning between $45,000$ and $65,000$ is $37.2%$

Let $x$ be the length of the component.

$x$ has $ \mu=5, \sigma=0.02$

We need to find the probability of the length of the component between $4.96$ and $5.04$. That is to find $P(4.96<x<5.04)$.

Given $x,z=\dfrac{x-\mu}{\sigma}$

Thus for $x=4.96,z=\dfrac{4.96-5}{0.02}=-2$

and for $x=5.04,z=\dfrac{5.04-5}{0.02}=2$

Therefore $P(4.96<x<5.04)=P(-2<z<2)$

                                                       $=P(z<2)-P(z<-2)$

                                                       $=0.9772-0.0228$ (from normal distribution table)

                                                       $=0.9544$

$ \therefore P(4.96<x<5.04)=0.9544$

Hence the probability of the length of the component between $4.96$ and $5.04$ is $0.9544$

Let $x$ be the random variable that represents the speed of cars.
$x$ has $\mu=90,\sigma=10$.

We have to find the probability that $x$ is higher than $100$ or $P(x > 100)$

Given $x, z=\dfrac{x-\mu}{\sigma}$.

Thus for $x=100, z=\dfrac{100-90}{10}=1$

$ \Rightarrow P(x>100)=P(z=1)$

                         $=$ [total area]$-$[area to the left of $z=1$]

                         $=1-0.8413$ (from normal distribution table)

$\therefore P(x>100)=0.1587$

Hence the probability that a car selected at a random has a speed greater than $100$ km/hr is equal to $0.1587$.