Tag: measures of dispersion and skewness

Questions Related to measures of dispersion and skewness

For a distribution, coefficient of variation is 22.5% and mean is 7.5 find the standard deviation___.

  1. 1.9975

  2. 1.6875

  3. 1.243

  4. 1.0943


Correct Option: B

In a class of 100, the mean on a certain exam was 50, the standard deviation, 0. This means..........................

  1. half the class had scores less than 50

  2. there was a high correlation between ability and grade

  3. everyone had a score of exactly 50

  4. half the class had 0's and half had 50's


Correct Option: C
Explanation:

Standard deviation is the square root of the arithmetic mean of the squares of the deviations measured from the arithmetic mean of the data. If standard deviation of a certain series is zero than it denotes that all the values of that series is equal to the mean of the series which made all the deviations zero and therefore standard deviation also zero. 

Average Marks of a group of $50$ students appeared in CA CPT exams is $134$ marks. If $10\%$ students scored more than $137$ marks, find the standard deviation of marks_____.

  1. $1.98$

  2. $2.336$

  3. $3.05$

  4. $2.98$


Correct Option: B

If a population has a standard deviation of $2.25$, how large the sample should be to allow a maximum error of $0.33$ with $95\%$ confidence level?

  1. $252$

  2. $201$

  3. $220$

  4. $178$


Correct Option: D

The standard deviation $\sigma$ of the first $N$ natural numbers can be obtained using which one of the following formula?

  1. $\sigma =\cfrac { { N }^{ 2 }-1 }{ 12 } $

  2. $\sigma =\sqrt { \cfrac { { N }^{ 2 }-1 }{ 12 } } $

  3. $\sigma =\sqrt { \cfrac { N-1 }{ 12 } } $

  4. $\sigma =\sqrt { \cfrac { { N }^{ 2 }-1 }{ 6N } } $


Correct Option: C
Explanation:

$\sigma^2=\dfrac{1}{n}\sum _{i=1}^{n}i^2-\left(\dfrac{1}{n}\sum _{i=1}^{n}i\right)^2$


     $=\dfrac{1}{n}(1^2+2^2+...+n^2)-\left(\dfrac{1}{n}(1+2+...+n)\right)^2$

     $=\dfrac{1}{n} \times \dfrac{n(n+1)(2n+1)}{6}-(\dfrac{n+1}{2})^2=\dfrac{n^2-1}{12}$

$\therefore \sigma=\sqrt{\dfrac{n^2-1}{12}}$

A researcher has collected the following sample data. The mean of the sample is 5.
3  5  12  3  2
The standard deviation is...............

  1. 8.944

  2. 4.062

  3. 13.2

  4. 16.5


Correct Option: B