The mean of the following frequency distribution is 35. Find the values of x and y, if the sum of frequencies is 25 :

Class: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70

Frequency: 1, x, 5, 7, y, 3, 1

Q: The mean of the following frequency distribution is 35. Find the values of x and y, if the sum of frequencies is 25 : Class: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70 Frequency: 1, x, 5, 7, y, 3, 1

$x = 3$ and $y = 5$, obtained by solving $x + y = 8$ and $x + 3y = 18$ together (from the frequency-sum and mean conditions respectively).

Statistics

Mean of Grouped Frequency Distribution

Official Answer

$x = 3$ and $y = 5$ , obtained by solving $x + y = 8$ and $x + 3y = 18$ together (from the frequency-sum and mean conditions respectively).

mean of grouped dataclass markdirect methodfrequency distributionsimultaneous linear equationssum of frequenciesmissing frequency problem

Marking Scheme

11 mark: correctly writing the class marks (mid-points) 5, 15, 25, 35, 45, 55, 65.
21 mark: correctly forming the equation $x + y = 8$ from the frequency-sum condition.
31 mark: correctly computing $\Sigma f_i \cdot x_i = 605 + 15x + 45y$ .
41 mark: using the mean condition to form the second equation $x + 3y = 18$ (or equivalent $15x + 45y = 270$ ).
51 mark: correctly solving the simultaneous equations to get $x = 3$ , $y = 5$ , with verification.

Hint

Use $\Sigma f_i = 25$ to get one linear equation in x, y; use $\text{Mean} = \frac{\Sigma f_i \cdot x_i}{\Sigma f_i} = 35$ to get a second linear equation; solve the two equations simultaneously.

Quick Oral Answer

Using $\Sigma f_i = 25$ , we get $x + y = 8$ ; using the mean formula $\frac{\Sigma f_i \cdot x_i}{\Sigma f_i} = 35$ , we get $605 + 15x + 45y = 875$ , i.e., $x + 3y = 18$ ; solving these two equations together gives $y = 5$ and $x = 3$ .

Analysis & Explanation

A two-unknown statistics problem combining the direct-method mean formula with a pair of simultaneous linear equations.

Method

Compute class marks (mid-values) first: 5, 15, 25, 35, 45, 55, 65.
Frequency-sum condition gives $x + y = 8$ ; the mean condition ( $\text{Mean} = \frac{\Sigma f_i x_i}{\Sigma f_i}$ ) gives a second equation $x + 3y = 18$ .
Solve by elimination (subtract one equation from the other) to isolate y quickly.

Common mistakes

Arithmetic slips while computing class marks or multiplying frequencies by large mid-values (e.g., 7×35, y×45) — keeping a running table avoids this.
Substituting instead of eliminating, which is slower here since the coefficients differ only in y.

Real-world connection

This "two missing frequencies" technique mirrors how statisticians estimate missing survey/census cells from partial data and a known summary statistic like the mean.

Common Mistakes

1Computing class marks incorrectly (e.g., using the upper limit alone instead of the average of lower and upper limits).
2Arithmetic errors while expanding $\Sigma f_i \cdot x_i$ , especially multiplying larger frequencies like 7 and unknowns x, y by their respective class marks.
3Setting up the mean equation incorrectly by forgetting to divide by the total frequency (25) before equating to 35.

Interesting Facts

The 'find missing frequencies given the mean' problem type is one of the most repeated CBSE Statistics questions, appearing in nearly every board exam year in some variant (2 unknowns with mean, or 1 unknown with mean/median/mode).

The direct method used here is one of three standard methods (Direct, Assumed Mean, Step-Deviation) taught in NCERT for computing the mean of grouped data — the step-deviation method is often faster when class marks are large.

This exact technique — solving simultaneous equations from summary statistics — is a simplified version of how real census and survey agencies estimate unrecorded frequency counts when only aggregate statistics are published.

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Frequently Asked Questions

Why do we need two equations to find x and y?

Since there are two unknown frequencies (x and y), one equation alone (from the frequency sum) is not enough to find unique values; the mean condition provides the second independent equation needed to solve for both unknowns.

Could the step-deviation method be used instead of the direct method here?

Yes, the step-deviation method would also work and might involve smaller numbers, but since the class marks here are not very large, the direct method is simpler and equally efficient for this particular problem.

How can the final answer be verified quickly?

Substitute $x = 3$ and $y = 5$ back into the frequency list, confirm the frequencies sum to 25, then recompute $\Sigma f_i \cdot x_i$ and divide by 25 to confirm the mean equals 35.