Tag: introduction to normal distribution

Questions Related to introduction to normal distribution

The average length of time required to complete a jury questionnaire is $40$ minutes, with a standard deviation of $5$ minutes. What is the probability that it will take a prospective juror between $30$ and $50$ minutes to complete the questionnaire?

statistics normal distribution distribution of measurement probability distributions introduction to normal distribution

  1. About $85\%$

  2. About $90\%$

  3. About $95\%$

  4. none

Show answer
Correct Option: C
Explanation:
For normal distribution $ P(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/(2\sigma^2)}$
Average length of time required to complete a jury questionnaire is 40 minutes
Mean ($ \mu $)=40
standard deviation($\sigma $)=5
Probability that it will take 30 to 50 min i.e, P(30<x<50)
For standard normal distribution curve
Z=(x-$ \mu $)/$\sigma $
(30-40)/5<(x-$ \mu $)/$\sigma $<(50-40)/5
-2<Z<2

For standard normal distribution mean shifted to zero, $ P(x)dx=\frac{1}{\sqrt{2\pi}}e^{-Z^2/2}dz$ 
Also,P(-2<Z<2)=area of the region between -1 to 1 that is approximately equal to=95%

The marks secured by $400$ students in a Mathematics test were normally distributed with mean $65$. If $120$ students got marks above $85$, the number of students securing marks between $45$ and $65$ is

statistics normal distribution distribution of measurement probability distributions introduction to normal distribution

  1. $120$

  2. $20$

  3. $80$

  4. $160$

Show answer
Correct Option: C
Explanation:

Let $X$ denote the marks secured.

Given, $\mu =65$
Thus, $X\sim N(65,\rho)$
$\Rightarrow z=\dfrac {X-\mu}{\rho}=\dfrac {X-65}{\rho}$
$\Rightarrow P(X>85)=\dfrac {120}{400}$
$\Rightarrow P\left (z>\dfrac {85-65}{\rho}\right)=\dfrac {3}{10}$
$\Rightarrow P\left (z>\dfrac {20}{\rho}\right)=\dfrac {3}{10}$ ....(1)
$\Rightarrow P(45<x<65)$ $=P\left (\dfrac {45-65}{\rho}<z<\dfrac {65-65}{\rho}\right)$
$=P\left (\dfrac {-20}{\rho}<z<0\right)$
$=P\left (0<z<\dfrac {20}{\rho}\right)$
$=0.5-P\left (z>\dfrac {20}{\rho}\right)$
$=\dfrac {1}{2}-\dfrac {3}{10}$
$=\dfrac {1}{5}$
Number of students secured marks between $45$ and $65$ $=\dfrac {1}{5}\times 400=80$.
Hence, the correct answer is option .

The annual salaries of employees in a large company are approximately normally distributed with a mean of $50,000$ and a standard deviation of $20,000$.  What percent of people earn between $45,000$ and $65,000$?

statistics normal distribution distribution of measurement probability distributions introduction to normal distribution

  1. $56.23$%

  2. $47.4$%

  3. $37.2$%

  4. $38.56$%

Show answer
Correct Option: C
Explanation:

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 $37.2\%$

The length of similar components produced by a company is approximated by a normal distribution model with a mean of $5$ cm and a standard deviation of $0.02$ cm. If a component is chosen at random, what is the probability that the length of this component is between $4.96$ and $5.04$ cm?

statistics normal distribution distribution of measurement probability distributions introduction to normal distribution

  1. $0.9544$

  2. $0.1236$

  3. $0.7265$

  4. $0.9546$

Show answer
Correct Option: A
Explanation:

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 $0.9544$

A radar unit is used to measure speeds of cars on a motorway. The speeds are normally distributed with a mean of $9$ km/hr and a standard deviation of $10$ km/hr. What is the probability that a car picked at random is travelling at more than $100$ km/hr?

statistics normal distribution distribution of measurement probability distributions introduction to normal distribution

  1. $0.1698$

  2. $0.1548$

  3. $0.1587$

  4. $0.1236$

Show answer
Correct Option: C
Explanation:

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 $0.1587$.