### Distribution of measurement - class-XI

 Description: distribution of measurement Number of Questions: 15 Created by: Rachana Sahu Tags: statistics measurements and uncertainties physics maths probability distributions normal distribution probability
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$\sum _{r=1}^{11} r.5^{r} =\dfrac{(43\times 5^{a}+5)}{b}$, 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 $\left(\sigma\right)$.

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:
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 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. $160$

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 $65,000$?

1. $56.23$%

2. $47.4$%

3. $37.2$%

4. $38.56$%

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?

1. $0.9544$

2. $0.1236$

3. $0.7265$

4. $0.9546$

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?

1. $0.1698$

2. $0.1548$

3. $0.1587$

4. $0.1236$

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 $0.1587$.

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