Tag: statistics

Quizzes (51) Questions (221)

Questions Related to statistics

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

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

  1. $120$
  2. $20$
  3. $80$
  4. $160$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 .

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

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 $65,000$?

  1. $56.23$%
  2. $47.4$%
  3. $37.2$%
  4. $38.56$%
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $37.2\%$

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

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?

  1. $0.9544$
  2. $0.1236$
  3. $0.7265$
  4. $0.9546$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $0.9544$

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

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?

  1. $0.1698$
  2. $0.1548$
  3. $0.1587$
  4. $0.1236$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $0.1587$.

Multiple choice statistics skewness of frequency distribution karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

Karl Pearson's coefficient of skewness of a distribution is 0.32.Its s.d.is 6.5 and mean is 29.6.The mode and median of the distribution are

  1. 27.52,28.91

  2. 26.92,27.23

  3. 25.67,26.34

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Karl Pearson's coefficient of skewness $\displaystyle =\frac{Mean-Mode}{S.D.}$ $\displaystyle \therefore 0.32=\frac{29.6-Mode}{6.5}\Rightarrow Mode=27.52$ Also Karl Pearson's coeff.of skewness $\displaystyle =\frac{3\left ( Mean-Median \right )}{S.D}$ $\displaystyle \because 0.32=\frac{3\left ( 29.6-Median \right )}{6.5}$ $\displaystyle \Rightarrow Median=28.91$

Multiple choice statistics skewness of frequency distribution karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

The sum of the deviations of the variates 6,8,10,16,20,24 

  1. -1

  2. 1

  3. 0

  4. 4
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\Rightarrow$   $Mean = \dfrac{6+8+10+16+20+24}{6}=14$

$\Rightarrow$   Sum of the deviations = $(6-14)+(8-14)+(10-14)+(16-14)+(20-14)+(24-14)$
$\Rightarrow$   Sum of the deviation = $-8-6-4+2+6+10$
$\therefore$    Sum of the deviation = $-18+18$
$\therefore$    Sum of the deviation = $0$

Multiple choice statistics skewness of frequency distribution karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

Choose the statement which consists of two correlated variables.

  1. Increase in the intensity of cold results in greater sale of woollen clothes

  2. Increase in temperature of delhi has led to congestion

  3. Increase in weight of children s accompanied by increase in weight of their mother

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since increase in intensity of cold result in greater scale of woolen clothes,

Multiple choice statistics skewness of frequency distribution karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

Calculate Pearson's coefficient of correlation between the values of $X$ and $Y$.

X 1 2 3 4 5
Y 7 6 5 4 3
  1. $-0.3$
  2. $0.3$
  3. $1$
  4. $-1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$X\\ 1\\ 2\\ 3\\ 4\\ 5\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \overline { x } \cfrac { 15 }{ 5 } =3$                $Y\\ 7\\ 6\\ 5\\ 4\\ 3\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \overline { y } =\cfrac { 25 }{ 5 } =5$                  $Y=y-\overline { y } \\ \quad 02\\ \quad 01\\ \quad 00\\ -1\\ -1\\ \ _ \ _ \ _ \ _ \ _ \ _ \\ \quad 0$                    $XY\\ -4\\ -1\\ \quad 0\\ -1\\ -4\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \sum { XY=-10 } $                 ${ X }^{ 2 }\\ 4\\ 1\\ 0\\ 1\\ 4\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \sum { { X }^{ 2 }=10 } $        

${ Y }^{ 2 }\\ 4\\ 1\\ 0\\ 1\\ 4\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \sum { { Y }^{ 2 }=10 } $

Therefore, $r=\cfrac { \sum { XY }  }{ \sqrt { \sum { { X }^{ 2 }\sum { { Y }^{ 2 } }  }  }  } \\ =\cfrac { -10 }{ \sqrt { 10*10 }  } \\ =\cfrac { -10 }{ 10 } \\ =-1.$
Multiple choice statistics skewness of frequency distribution karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

For $ n=25,\sum x=125,\sum x^2=650,\sum y=100,\sum y^2=460,\sum xy=508$, correlation coefficient is 

  1. $0.99$
  2. $0.207$
  3. $0.66$
  4. $0.89$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$n=25$

$\sum x=125,\ \sum{{x}^{2}}=650,\ \sum{xy}=334$  $\sum y=100,\ \sum{{y}^{2}}=508$
$r=\cfrac { n\sum { xy } -(\sum { x } \times \sum { y } ) }{ \sqrt { (n\sum { { x }^{ 2 } } -\sum { { x }^{ 2 } } )(n\sum { { y }^{ 2 } } -\sum { { y }^{ 2 } } ) }  } =\cfrac { 25\times 508 -100\times 125 }{ \sqrt { (25\times 650 -{ 125 }^{ 2 }  )(25\times 460 -{100}^{2}) }  }=0.207 $