• edited by
2,430 views
2 2 votes

A class of first year B.tech students is composed of four batches A, B, C and D, each consisting of $30$ students. It is found that the sessional marks of students in Engineering Drawing in batch C have a mean of $6.6$ and standard deviation of $2.3$. The mean and the standard deviation of the marks for the entire class are $5.5$ and $4.2$ respectively. It is decided by the course instructor to normalize the marks of the students of all batches to have the same mean and standard deviation as that of the entire class. Due to this, the marks of a student in batch C are changed from $8.5$ to

  1. $8.75$
  2. $7.45$
  3. $9.27$
  4. $8.97$

3 Answers

Best answer
15 15 votes

Mean of 30 marks must be reduced from 6.6 to 5.5 while SD must increase from 2.3 to 4.2. 

Multiplying each number by n changes SD by n (Since SD = $\sum_i{(x_i-\mu)}^2$). So, we can make the SD change from 2.3 to 4.2 by multiplying each term by 4.2/2.3. This will also change the mean from 6.6 to 12.05. 

Now, subtracting a constant from each term makes the mean reduce by that same constant. (since $\mu = \frac{\sum_i{x_i}}{n}$). So, to reduce the mean from 12.05 to 5.5 we should subtract 6.55 from each term. (Subtracting a constant from each term won't affect the standard deviation as mean and each term change by the same amount and hence also won't change).

So, our given value 8.5 becomes 8.5 * 4.2 / 2.3 = 15.52 for adjusting SD, and then becomes 15.52 - 6.55 = 8.97 when adjusted for mean. 

• selected by
1 1 vote
Mean of 30 marks must be reduced from 6.6 to 5.5 while SD must increase from 2.3 to 4.2.

Multiplying each number by n changes SD by n (Since SD = ∑i(xi−μ)2∑i(xi−μ)2). So, we can make the SD change from 2.3 to 4.2 by multiplying each term by 4.2/2.3. This will also change the mean from 6.6 to 12.05.

Now, subtracting a constant from each term makes the mean reduce by that same constant. (since μ=∑ixi/n). So, to reduce the mean from 12.05 to 5.5 we should subtract 6.55 from each term. (Subtracting a constant from each term won't affect the standard deviation as mean and each term change by the same amount and hence also won't change).

So, our given value 8.5 becomes 8.5 * 4.2 / 2.3 = 15.52 for adjusting SD, and then becomes 15.52 - 6.55 = 8.97 when adjusted for mean.
Answer:
Position:
Show:

Related questions

4 4 votes
2 answers 2 answers
1.7k
1.7k views
Ruturaj Mohanty asked Dec 27, 2018
1,656 views
For two data sets, each of size $5$, the variances are given to be $4$ and $5$ and the corresponding means are given to be $2$ and $4$ respectively. The variance of the c...
1 1 vote
1 answers 1 answer
2.0k
2.0k views
Ruturaj Mohanty asked Dec 27, 2018
2,042 views
What is the minimum number of people that must be there in a room to make the probability of two people having same birthday be at least 50%? Assume a year has $365$ day...
4 4 votes
2 answers 2 answers
4.3k
4.3k views
Ruturaj Mohanty asked Dec 27, 2018
4,344 views
A book contains $100$ pages. A page is chosen at random. What is the chance that the sum of the digits on the page is equal to $8$?$0.08$$0.09$$0.90$$0.10$
6 6 votes
2 2 answers
2.4k
2.4k views
Ruturaj Mohanty asked Dec 27, 2018
2,374 views
$\begin{bmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix}$For the above given matrix $A,$$A^3 -7A^2 +10A = $$5I+A$$5I-A$$A-5I$$6I$