Mathematics

Probability Distributions

457 Questions

Probability distributions describe the likelihood of different outcomes in a statistical experiment. Key topics include calculating confidence intervals, understanding Type II errors, and working with the standard normal distribution. These concepts are frequently tested in mathematics and statistics sections of various competitive exams.

Standard normal distributionType II error probabilityConfidence interval interpretationPoisson distributionProbability density function

Probability Distributions Questions

Multiple choice business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions


 lf the mean is $\lambda$ and the variance is $\sigma^{2}$ in a Poisson distribution, then

  1. $\displaystyle \lambda=\frac{1}{2}\sigma^{2}$
  2. $\displaystyle \sigma^{2}=\frac{1}{2}\lambda$
  3. $\lambda=\sigma^{2}$
  4. $\sigma^{2}=\lambda^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a Poisson distribution, mean and variance are equal .
i.e.$\lambda = \sigma^2$

Multiple choice business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

If ${\overline{x}}$ and $\sigma^{2}$ are mean and variance of poisson distribution, then

  1. $\overline{x}>\sigma^{2}$
  2. $\overline{x}<\sigma^{2}$
  3. $\overline{x}=\sigma^{2}$
  4. $\overline{x}+\sigma^{2}=1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a Poisson distribution, mean ($\mu$) and variance ($\sigma^2$ )are equal .
i.e.$\mu = \sigma^2$

Multiple choice business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

A examinations consists of $8$ questions in each of which  one of the $5$ alternatives is the correct one. On the assumption that a candidate who has done no preparatory  work, chooses for each questions any one of the five alternatives with equal probability, then the probability that he gets more than one correct answer is equal to:

  1. ${\left( {0.8} \right)^8}$
  2. $3{\left( {0.8} \right)^8}$
  3. $1-{\left( {0.8} \right)^8}$
  4. $1-3{\left( {0.8} \right)^8}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Probability of an answers to be correct $=\dfrac{1}{5}=0.2$

Probability of an answers not to be correct $=1-0.2$
$=0.8$

Probability $\left(more\ than\ 1\ correct\right)=1-P\left(0\ correct\right)-P\left(1\ correct\right)$

$=1-^{8}{C} _{0}{\left(0.8\right)}^{8}-^{8}{C} _{1}\left(0.2\right) \left(0.8\right)$

$=1-{\left(0.8\right)}^{8}-1.6{\left(0.8\right)}^{7}$

$=1-{\left(0.8\right)}^{8}-2{\left(0.8\right)}^{8}$

$=1-3{\left(0.8\right)}^{8}$

$D$ is coorect.
Multiple choice business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

For a poission distribution variable $X$ is such that $P(X = 2) = 9 P(X= 4) + 90 P(X= 6)$ the mean is

  1. $2$
  2. $3$
  3. $1$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
For $P.D. P(X=r)=\displaystyle \dfrac{e^{-\lambda}\lambda ^{r}}{r!},r=0,1,2,\cdots$
$\therefore \displaystyle \dfrac{e^{-\lambda}\lambda ^{2}}{2!}=\dfrac{9e^{-\lambda}\lambda ^{4}}{4!}+90 \dfrac{e^{-\lambda}\lambda ^{6}}{6!}$        (given)
$\Rightarrow \displaystyle\dfrac{\lambda ^{2}}{2}=\dfrac{9}{24}\lambda ^{4}+\dfrac{90}{720}\lambda ^{6}$ 
$\Rightarrow \lambda ^{4}+3\lambda ^{2}-4=0$
$\Rightarrow \left ( \lambda^{2}+4 \right )\left ( \lambda^{2}-1 \right )=0=>\lambda= \pm 1$
$\Rightarrow \lambda=1$ as $\lambda>0$ and $(\lambda^{2}+4=0$ impossible$)$
$\therefore$ mean$=\lambda=1$
Multiple choice business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

For a Poission distribution, which of the following is true

  1. $Mean = Mode$
  2. $Median = S.D.$
  3. $Mean = Variance$
  4. $Median = Variance$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
For Poission distribution we have $\displaystyle P\left ( r \right )=\dfrac{e^{-\pi }\lambda ^{r}}{r^{i}}\left ( r=0, 1, 2,...\infty  \right)$ 
$\displaystyle$ mean $\displaystyle =\dfrac{\sum f _{i}x _{i}}{\sum f _{i}}=\sum _{r=0}^{\infty }rP\left ( r \right ),\sum f _{i}P\left ( r\right )=1$$\displaystyle =0+\lambda e^{-\lambda }+2\dfrac{\lambda ^{2}e^{-\lambda }}{2!}+3\dfrac{\lambda ^{3}e^{-\lambda }}{3!}+.....\infty$ $\displaystyle =\lambda e^{-\lambda }\left ( 1+\dfrac{\lambda }{11}+\dfrac{\lambda ^{2}}{2!}+\dfrac{\lambda ^{3}}{3!}+.....\infty  \right )$$\displaystyle =\lambda e^{\lambda }.e^{\lambda }=\lambda $
similarly, $\displaystyle o^{2}=$ Variance $=\sum _{r=0}^{\infty }r^{2}P\left ( r \right )-\left ( \sum _{r=0}^{\infty } rP\left ( r \right ) \right )^{2}$ $=\displaystyle \lambda e^{\lambda }\left ( e^{-\lambda }+\lambda e^{-\lambda } \right )-\lambda ^{2}=\lambda =$ mean 
$\therefore \text{mean}=\text{variance}$