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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. $8.97$

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.

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E[aX+b] = aE[X] + b

Var[aX+b] = (a^2) Var[X]

SD[aX] = (Var[aX])^(1/2) = a * SD[X]

Multiplying SD by a changes SD and mean both by a.

Mean change from 6.6 to 12.05 as multiplied by 4.2/2.3

SD measures spread so, adding constant to all values doesn’t change SD.

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.
X – 5.5 / 4.2  =  8.5 – 6.6 / 2.3  = 8.9

Which is equivlaent to 9

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