Tag: distribution of measurement

Questions Related to distribution of measurement

$\sum _{r=1}^{11} r.5^{r} =\dfrac{(43\times 5^{a}+5)}{b}$, then $(a+b)$ is

  1. $18$

  2. $28$

  3. $15$

  4. $38$


Correct Option: A

If $A$ and $B$ are two independent events such that $P(A) = \dfrac{1}{2}$ and $P(B) = \dfrac{2}{3}$, then $P((A \cup B) (A\cup \overline{B})(\overline{A} \cup B))$ has the value equal to 

  1. $\dfrac{1}{3}$

  2. $\dfrac{1}{4}$

  3. $\dfrac{1}{2}$

  4. $\dfrac{2}{3}$


Correct Option: A

Which of the following is not true regarding the normal distribution?

  1. the point of inflecting are at $X = \mu \pm \sigma$

  2. skewness is zero

  3. maximum heigth of the curve is $\dfrac{1}{\sqrt{2\pi}}$

  4. mean $=$ media $=$ mode


Correct Option: C
Explanation:
$\left(i\right)$Since $f\left(x\right)$ is a nonzero function we may divide both sides of the equation by this function. From this 
it is  easy to see that the inflection points occur where $X =\mu\pm\sigma$. In other words the inflection points are 
located one standard deviation above the mean and one standard deviation below the mean
$\left(ii\right)$The skewness for perfect normal distribution is $0.0$. But if sample is greater than $100$ and less 
than $200$, the acceptable absolute skewness value is $1.0$. However for large sample size $n$ greater than $200$, 
the absolute value for acceptable skewness is $1.5$.
$\left(iii\right)$The area under the normal curve is equal to $1.0$. Normal distributions are denser in the center and
 less dense in the tails. Normal distributions are defined by two parameters, the mean $\left(\mu\right)$ and the standard
deviation $\left(\sigma\right)$. $68\%$ of the area  of a normal distribution is within one standard deviation of the mean.
$\left(iv\right)$The mean, median, and mode of a normal distribution are equal. The area under the normal curve is equal to 1.0. 
Normal distributions are denser in the center and less dense in the tails. Normal distributions are defined by two 
parameters, the mean $\left(\mu\right)$ and the standard deviation $\left(\sigma\right)$.

The random variable $x$ follows normal distribution $f(x) = Ce^{\dfrac{-\dfrac{1}{2} (x - 100)^2}{25}}$. Then the value of $C$ is

  1. $\sqrt{2\pi}$

  2. $\dfrac{1}{\sqrt{2 \pi}}$

  3. $5\sqrt{2 \pi}$

  4. $\dfrac{1}{5 \sqrt{2 \pi}}$


Correct Option: D
Explanation:

Since the above variable $x$ follows normal distribution, we can therefore write it as 


$Ce^{-\dfrac{(x-100)^{2}}{50}}=\dfrac{1}{\sqrt{2\sigma^{2}\pi}}e^{-\dfrac{(x-\mu)^{2}}{2\sigma^{2}}}$

Comparing LHS with RHS gives 

$2\sigma^{2}=50$...(i)

And $C=\dfrac{1}{\sqrt{2\sigma^{2}\pi}}$...(ii)

Now from equation we know that $2\sigma^{2}=50$. Substituting in eq (i) gives

$C=\dfrac{1}{\sqrt{2\sigma^{2}\pi}}=\dfrac{1}{\sqrt{50\pi}}$

$=\dfrac{1}{5\sqrt{2\pi}}$

Which of the following are correct regarding normal distribution curve?
(i) Symmetrical about the line $X=\mu $ (Mean)
(ii) Mean $=$ Median $=$ Mode
(iii) Unimodal
(iv) Points of inflexion are at $X=\mu \pm \sigma $

  1. (i), (ii)

  2. (ii), (iv)

  3. (i), (ii), (iii)

  4. All of these


Correct Option: D
Explanation:

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

$x=$ mean (mode or median).
It is unimodal as it has its first derivative $0$ at $x=\mu$ (mean) and the derivative is less than $0$, for $x>\mu$ and greater than $0$ for $x<\mu$
It has its point of inflection (second derivative $0$) at $x=\mu+\sigma$ and $x=\mu-\sigma$ where $\sigma$ is the standard deviation.

$X$ is a Normally distributed variable with mean $ = 30$ and standard deviation $ = 4$. Find $P(30 < x<35)$

  1. $0.3698$

  2. $0.3956$

  3. $0.2134$

  4. $0.3944$


Correct Option: D
Explanation:
Here we need to determine the value of $ Z=\dfrac{x-\mu}{\sigma} $
Here, $\mu=\text{Mean}=30$ and $\sigma=\text{Stand Deviation}=4$
For, $x=30\Rightarrow Z=\dfrac{30-30}{4}=0$
For, $x=35\Rightarrow Z=\dfrac{35-30}{4}=1.25$
$\Rightarrow P(30<x<35)=P(0<Z<1.25) $
$\Rightarrow P(0<Z<1.25)=P(Z<1.25)-P(Z<0) $                         $(\because P(a<Z<b)=P(Z<b)-P(Z<a))$
From the Normal Distribution table, $P(Z<1.25)=0.8944$ and $P(Z<0)=0.5$
$\Rightarrow P(30<x<35)=0.8944-0.5=0.3944$

A large group of students took a test in Physics and the final grades have a mean of $70$ and a standard deviation of $10$. If we can approximate the distribution of these grades by a normal distribution, what percent of the students should fail the test (grades$<60$)?

  1. $15.21$

  2. $23.21$

  3. $15.87$%

  4. $16.23$


Correct Option: C
Explanation:
Final grades follows normal distribution.
Given mean i.e. $ \mu $ $= 70$
Given standard deviation i.e. $ \sigma $ $= 10$

The normal random variable of a standard normal distribution is called a standard score or a z-score. Every normal random variable X can be transformed into a z score via the following equation:

$z = (X - μ) / σ$

where $X$ is a normal random variable, $μ$ is the mean of $X$, and $σ$ is the standard deviation of $X$.


For $X =60$
$Z= (60-70)/10 = -1$

$P(X < 60) = P(Z < -1)$ 
                $= 0.1587$
Hence percent of students failed in test is $15.87\%$

The length of life of an instrument produced by a machine has a normal distribution with a mean of $12$ months and standard deviation of $2$ months. Find the probability that an instrument produced by this machine will last less than $7$ months. 

  1. $0.2316$

  2. $0.0062$

  3. $0.0072$

  4. $0.2136$


Correct Option: B
Explanation:
Life of intsrument follows normal distribution.
Given mean i.e. $ \mu $ $= 12 $months
Given standard deviation i.e. $ \sigma $ $= 2$ months

The normal random variable of a standard normal distribution is called a standard score or a z-score. Every normal random variable X can be transformed into a z score via the following equation:

$z = (X - μ) / σ$

where $X$ is a normal random variable, $μ$ is the mean of $X$, and $σ$ is the standard deviation of $X$.

For $X =22$
$Z= (7-12)/2 = -2.5$

$P( X < 7) = P(Z < -2.5)$
              $ = 0.0062$

The scores on standardized admissions test are normally distributed with a mean of $500$ and a standard deviation of $100$. What is the probability that a randomly selected student will score between $400$ and $600$ on the test?

  1. About $63\%$

  2. About $65\%$

  3. About $68\%$

  4. About $70\%$


Correct Option: C
Explanation:
For normal distribution $ P(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/(2\sigma^2)}$
The scores on standardised admissions test are normally distributed with a mean of 500

Mean ($ \mu $)=500

standard deviation($\sigma $)=100
Probability that score lies between 400 and 600  i.e, P(400<x<600)
For standard normal distribution curve
Z=(x-$ \mu $)/$\sigma $
(400-500)/100<(x-$ \mu $)/$\sigma $<(600-500)/100
-1<Z<1
For standard normal distribution mean shifted to zero, $ P(x)dx=\frac{1}{\sqrt{2\pi}}e^{-Z^2/2}dz$ 
Also,P(-1<Z<1)=area of the region between -1 to 1 that is approximately equal to=68%

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 $35$ and $45$ minutes to complete the questionnaire?

  1. About $68\%$

  2. About $72\%$

  3. About $76\%$

  4. About $84\%$


Correct Option: A
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 $)=5
Probability that it will take 35 to 45 min i.e, P(35<x<45)
For standard normal distribution curve
Z=(x-$ \mu $)/$\sigma $
(35-40)/5<(x-$ \mu $)/$\sigma $<(45-40)/5
-1<Z<1
For standard normal distribution mean shifted to zero, $ P(x)dx=\frac{1}{\sqrt{2\pi}}e^{-Z^2/2}dz$ 
Also,P(-1<Z<1)=area of the region between -1 to 1 that is approximately equal to=68%