Question 11 Mark
Mean of a normal distribution is $13.25$ and its standard deviation is $10.$ Estimate the value of its third quartile.
AnswerIn a normal distribution $Q_3=\mu+0.675 \sigma$.
Therefore $Q_3=13.25+0.675(10)=13.25+6.75=20$.
Thus, estimate value of its third quartile is $20 .$
View full question & answer→Question 21 Mark
What percentage of area is covered under the normal curve within the range $\mu-2 \sigma$ to $\mu+2 \sigma ?$
AnswerPercentage of area is covered under the normal curve within the range $μ - 2σ$ to $μ + 2σ$ is $95.45.$
View full question & answer→Question 31 Mark
Which value of normal variable divides the area of normal curve in two equal parts?
AnswerNormal curve is symmetric on both the sides for mean of normal variable $X.$
Therefore for $X = \mu $ of normal variable divides the area of normal curve in two equal parts.
View full question & answer→Question 41 Mark
For which value of standard normal variable, the standard normal curve is symmetric on both the sides?
AnswerFor $Z = 0$ of standard normal variable the standard normal curve is symmetric on both the sides.
View full question & answer→Question 51 Mark
“Standard score is independent of unit of measurement”, Is this statement true or false?
Answer“Standard score is independent of unit of measurement.’ It is a true statement.
View full question & answer→Question 61 Mark
What is the skewness of normal distribution?
AnswerThe skewness of normal distribution is $‘0’ ($zero$).$
View full question & answer→Question 71 Mark
What is the shape of normal curve?
AnswerThe shape of normal curve is completely bell-shaped.
View full question & answer→Question 81 Mark
What is the probability that a continuous random variable takes definite value?
AnswerThe probability that a continuous random variable takes definite value is always $‘0’; ($zero$).$
View full question & answer→Question 91 Mark
Marks obtained by students of a school in Economics subject follow normal distribution with mean $\mu$ and standard deviation $\sigma$. The value of standard score that a randomly selected student obtained $60$ marks is $1 .$ If the variance of variable is $100 ($marks$) ^2$ then find average marks.
AnswerHere, $x=60 ; Z=1 ; \sigma^2=100$
$\therefore \sigma=10$
$\therefore Z=\frac{x-\mu}{\sigma}$
$ \therefore 1=\frac{60-\mu}{10}$
$ \therefore 10=60-\mu$
$ \therefore \mu=60-10$
$ \therefore \mu=50$
Hence, the average marks obtained is $50 .$
View full question & answer→Question 101 Mark
The age of a group of persons follows normal distribution with mean $45$ years and standard deviation $10$ years. Calculate $Z-$score for a randomly selected person having age $60$ years.
AnswerHere, $\mu=45 ; \sigma=10, x-60$
$\therefore Z=\frac{x-\mu}{\sigma}=\frac{60-45}{10}=1.5$
Hence, $Z-$score $=1.5$
View full question & answer→Question 111 Mark
The monthly expense of a group of persons follows normal distribution with mean $Rs. 10, 000$ and standard deviation $Rs. 1, 000.$ A student has obtained a $Z-$score $= Rs. 1$ for randomly selected person having monthly expense more than $Rs. 11, 000.$ Is this calculation of $Z-$score true? Give reason.
AnswerHere, $\mu=10,000 ; \sigma=1000$
For $x>11000, Z>\frac{x-\mu}{\sigma}>\frac{11000-10000}{1000}>1$
A student has obtained a $Z-$score $=1$.
Hence, this calculation of $Z-$score is not true.
Reason: For given data, we must have $Z \geq 1$.
View full question & answer→Question 121 Mark
The extreme quartile of normal distributions is $20$ and $30.$ Find its mean.
AnswerHere, $\mathrm{Q}_1=20 ; \mathrm{Q}_3=30$
$\therefore \mathrm{M}=\frac{Q_3+Q_1}{2}=\frac{30+20}{2}=\frac{50}{2}=25$
In normal distribution, Mean = Median = Mode
$\therefore$ Mean $\mu=25$
View full question & answer→Question 131 Mark
For a probability distribution of standard normal variable, state the estimated limits for the middle $50\%$ observation.
AnswerFor a probability distribution of standard normal variable, first quartile $\left(Q_1\right)=-0.675$ and third quartile $\left(Q_3\right) 2 0.675.$
Therefore estimated limits for the middle $50 \%$ observation is between $-0.675$ and $0.675 .$
View full question & answer→Question 141 Mark
For a normal distribution the estimated value of quartile deviation is $12.$ Find the value of its standard deviation.
AnswerHere, quartile deviation $=12$
In normal distribution,
Estimated value of quartile deviation $\approx \frac{2}{3} \sigma$
$\therefore 12 \approx \frac{2}{3} \sigma$
$ \therefore \sigma \approx \frac{12 \times 3}{2} \approx 18$
Hence, the value of standard deviation obtained is $18 .$
View full question & answer→Question 151 Mark
The approximate value of mean deviation for a normal distribution is $8.$ Find its standard deviation.
AnswerHere, mean deviation $=8$
In normal distribution, mean deviation $\approx \frac{4}{5} \sigma$
$\therefore 8 \approx \frac{4}{5} \sigma$
$ \therefore \sigma \approx \frac{8 \times 5}{4} \approx 10$
Hence, the standard deviation obtained is $10.$
View full question & answer→Question 161 Mark
For a normal distribution having mean $10$ and standard deviation $6,$ estimate the value of quartile deviation.
AnswerHere, $\mu=10 ; \sigma=6$
Estimated value of quartile deviation:
In normal distribution, Quartile deviation $\approx \frac{2}{3} \sigma$
Putting, $\sigma=6$
Quartile deviation $\approx \frac{2}{3} \approx=4$
Hence, the estimated value of quartile deviation obtained is $4 .$
View full question & answer→Question 171 Mark
Give the values of the constant used in probability density function of normal variable.
AnswerThe values of the constant used in probability density function of normal variable is $\pi = 3.1416$ and $e = 2.7183$
View full question & answer→Question 181 Mark
If $P[0 \leq Z \leq 1.8]=0.4641$, then what percentage of observations in normal distribution are less than $\mu+1.8 \sigma$ ?
View full question & answer→Question 191 Mark
If the probability density function of a normal distribution is $f(x)=\frac{1}{4 \sqrt{2 \pi}} e^{\frac{-(x-25)^{2}}{32}}$, state the parameters of the distribution.
View full question & answer→Question 201 Mark
If $P \left[Z \leq z_{1}\right]=0.6915$, find $P\left[0 \leq Z \leq z_{1}\right]$
View full question & answer→Question 211 Mark
A normal distribution is $N(\mu, 16)$ For $X=52, Z$-score is $1 ,$ find the mean of distribution.
View full question & answer→Question 221 Mark
In a normal distribution for $x=90, Z-s c o r e=-0.6$. If $\mu=102$, find $\sigma$.
View full question & answer→Question 231 Mark
Median of a normal variable $X$ is 48 . If $Q_{1}=26$, find the mode and third quartile of the distribution.
Answer$Q_{3}=70, M_{0}=48$
View full question & answer→Question 241 Mark
How is the distance between quartiles and median of normal distribution?
AnswerThe quartiles and median of normal distribution are equidistant.
View full question & answer→Question 251 Mark
State the Interval for $99 \%$ of observations of standard normal distribution?
AnswerThe interval for $99 \%$ of observations of standard normal distribution Is $-2.575 \leq$ $Z \leq 2.575 .$
View full question & answer→Question 261 Mark
What is the value of skewness in the standard normal distribution?
AnswerIn standard normal distribution, the value of skewness is zero $(0)$.
View full question & answer→Question 271 Mark
The extreme quartiles of normal distribution are $25$ and $49 ,$ then what will be the value of parameter $\sigma$ of distribution?
View full question & answer→Question 281 Mark
The variance for a normal distribution is $2.25$. What will be the mean deviation of distribution?
View full question & answer→Question 291 Mark
In normal distribution, how will mean deviation be obtained?
AnswerIn normal distribution, the formula for mean deviation will be obtained as follows:Mean deviation $=\frac{4}{5} \sigma$, where $\sigma=$ standard deviation.
View full question & answer→Question 301 Mark
In normal distribution, write the formula for estimated value of quartile deviation.
AnswerIn normal distribution, the formula for estimated value of quartile deviation is as follows:Quartile deviation $=\frac{2}{3} \sigma$, where $\sigma=$ standard deviation.
View full question & answer→Question 311 Mark
What is the value of variance of distribution of standard normal variable $Z?$
AnswerThe value of variance of distribution of standard normal variable $Z$ is one $(1).$
View full question & answer→Question 321 Mark
State the parameters of a standard normal distribution.
AnswerThe parameters of a standard normal distribution are : mean $=0$ and standard deviation $1 .$
View full question & answer→Question 331 Mark
What is interval for the value of standard deviation of normal distribution?
AnswerThe interval for the value of standard deviation of normal distribution is $0<\sigma<\infty$
View full question & answer→Question 341 Mark
State the interval of mean of normal distribution.
AnswerThe interval of mean of normal distribution is $-\infty$
View full question & answer→Question 351 Mark
Write the probability density function of a normal distribution.
AnswerThe probability density function of a normal distribution is as follows: $f(x)=\frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{3-\mu}{\sigma}\right)^{2}},-\infty$
View full question & answer→Question 361 Mark
What is the symbol of the distribution of normal variable $X ?$
AnswerThe symbol of the distribution of normal variable $X$ is $N\left(\mu, \sigma^{2}\right)$
View full question & answer→Question 371 Mark
How is the distribution of normal distribution?
AnswerThe distribution is symmetric of normal distribution.
View full question & answer→Question 381 Mark
If $P \left(Z \geq z_1\right)=0.1480$, then find the value of $P \left(0 \leq z \leq z_1\right)$.
Answer$P \left(Z \geq z_1\right)=0.1480$
$\therefore P \left(0 \leq z \leq z_1\right)= P (0 \leq z \leq \infty)- P \left(Z \geq z_1\right)$
$=0.5-0.1480$
$=0.3520$
Thus, $P \left(0 \leq z \leq z_1\right)=0.3520$
View full question & answer→Question 391 Mark
In a normal distribution, deviation is $2.5$ times than its mean. If Mean is $40,$ then find standard deviation of distribution.
Answer
| $\sigma^2 = 2.5 \mu $ and $\mu = 40$ |
| $\therefore \sigma^2 = 2.5 \times 40 = 100$ |
| $\therefore \sigma^2 = 100$ |
| $\therefore \sigma = 10$ |
| $\therefore $ Standard deviation of distribution is $10.$ |
View full question & answer→Question 401 Mark
A monthly income of a group person follows normal distribution with mean $Rs. 2800$ and standard deviation $Rs. 200.$ Find $Z-$score for randomly selected person having monthly income $Rs. 3000.$
Answer
| Here, $\mu = 2800; \sigma =200$ and $x =3000$ |
| $\therefore \mathrm{Z}=\frac{\mathrm{x}-\mu}{\sigma}$ |
| $\therefore \mathrm{Z}=\frac{3000-2800}{200}$ |
| $\therefore Z=\frac{200}{200}=1$ |
| $\therefore Z-$score is $1.$ |
View full question & answer→Question 411 Mark
In a normal distribution $5Q_1 =2Q_2$. If its mean is $25$ then find its first quartile.
AnswerIn a normal distribution Values of mean, median and mode are same and $Q _2= M$
$\therefore 5 Q_1=(25)$
$\therefore 5 Q_1=50$
$\therefore Q_1=\frac{50}{5}=10$
$\therefore Q_1=10$
View full question & answer→Question 421 Mark
Median of normal distribution is $45$. Find the sum of its extreme quartiles.
Answer$M=\frac{Q_3+Q_1}{2}$
$\therefore 45=\frac{Q_3+Q_1}{2}$
$\therefore Q_3+Q_1=45 \times 2$
$\therefore Q_3+Q_1=90$
View full question & answer→Question 431 Mark
In a normal distribution sum of $Q_1$ and $Q_3=80$, then find its mean.
Answer$M=\frac{Q_3+Q_1}{2}=\frac{80}{2}=40$
In a normal distribution Values of mean, median and mode are same therefore Mean $(\mu)=40$.
View full question & answer→Question 441 Mark
For a normal probability distribution Mean $=$ Variance. If standard deviation is $15$ then what is the mean of distribution?
Answer
| Mean $(μ) =$ Variance $(\sigma)^2$ |
| Here, $\sigma = 15$ |
| $\therefore \sigma^2 = (15)^2 = 225$ |
| $\therefore $ Value of mean $(\mu ) = 225.$ |
View full question & answer→Question 451 Mark
The extreme quartiles of normal distribution are $48$ and $62.$ Find its median.
Answer$M =\frac{Q_3+Q_1}{2}$
$\therefore M =\frac{64+42}{2}$
$\therefore M =\frac{110}{2}$
$\therefore M =55$
View full question & answer→Question 461 Mark
In a normal distribution $Q_1=25, Q_2=40$. Find the value of $Q_3$.
Answer$M =\frac{Q_3+Q_1}{2}$
$\therefore 40=\frac{Q_3+25}{2}$
$\therefore 40 \times 2= Q _3+25$
$\therefore 80= Q _3+25$
$\therefore Q _3=80-25$
$\therefore Q _3=55$
View full question & answer→Question 471 Mark
The median of a normal distribution is $45.$ Find mean and mode of distribution.
AnswerIn a normal distribution values of mean, median and mode are same. Therefore mean find mode $= 45$
View full question & answer→Question 481 Mark
The variance of a normal variate is half of its mean. If the mean of the distribution is $10,$ write the probability function of the distribution.
Answer$\mu = 10$
$\therefore \sigma = \frac{10}{2} = 5$
$\therefore f(x)=\frac{1}{5 \sqrt{2 \pi}} e^{-1 / 2\left(\frac{x-10}{5}\right)^2} ;-\infty$
View full question & answer→Question 491 Mark
$"P \left(x_1 \leq X \leq x_2\right)=-0.0145 ".$ Is this value possible? Give reason for your answer.
Answer$P\left(x_1 \leq X \leq x_2\right)=-0.0145$. This value is not possible. because the probability is always positive.
View full question & answer→Question 501 Mark
State the area between $Z=0 \pm 2.575$ in normal curve.
Answer$99.73\%$ of observations of normal distribution lies in the interval $z = ± 3.$
View full question & answer→