TS EAMCET · Maths · Statistics

If $M_{1}$ and $M_{2}$ are the mean deviations from mean and median of the first $15$ even integers then $M_{1} + M_{2} =$

A $\frac{112}{15}$
B $\frac{224}{15}$
C $\frac{56}{15}$
D $\frac{28}{15}$

Verified Solution

Answer & Solution

Correct Answer

(B) $\frac{224}{15}$

Step-by-step Solution

Detailed explanation

First 15 even integers are 2, 4, 6, 8, 10, 12, 14,16, 18, 20, 22, 24, 26, 28 and 30 Mean x¯=2+4+6+8+...3015=16 Median x~=15+12th term=8th term=16 M1= mean deviation about mean=∑i=1nxi-x¯n…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From TS EAMCET

Explore more questions on app

If M1 and M2 are the mean deviations from mean and median of the first 15 even integers then M1+M2=

Step-by-step Solution

If $M_{1}$ and $M_{2}$ are the mean deviations from mean and median of the first $15$ even integers then $M_{1} + M_{2} =$