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.

Statistics

Finding missing frequencies from the mean

Official Answer

$x = 3$ and $y = 5$ .

mean of grouped datamissing frequencyclass mark midpointsum of frequencies 25x = 3y = 5simultaneous equationsdirect method mean

Marking Scheme

11 mark: correctly writing the class marks (midpoints) 5, 15, 25, 35, 45, 55, 65.
21 mark: forming the first equation from total frequency: $x + y = 8$ .
31 mark: correctly computing $\Sigma fx = 605 + 15x + 45y$ .
41 mark: forming and simplifying the mean equation to $x + 3y = 18$ .
51 mark: solving the pair to get $x = 3$ and $y = 5$ (verification acceptable for the final mark).

Hint

Use class marks (5, 15, …, 65). Set up two equations: $\Sigma f = 25$ gives $x + y = 8$ , and $\text{Mean} = \frac{\Sigma fx}{\Sigma f} = 35$ gives $x + 3y = 18$ ; solve simultaneously.

Quick Oral Answer

Taking class marks 5 to 65, the total frequency 25 gives $x + y = 8$ and the mean 35 gives $\Sigma fx = 875$ leading to $x + 3y = 18$ ; solving the two gives $y = 5$ and $x = 3$ .

Analysis & Explanation

A grouped-data statistics problem requiring the direct method for the mean together with simultaneous linear equations.

Concept

Class marks (midpoints) for equal-width classes are 5, 15, 25, 35, 45, 55, 65.
Total frequency condition: $1 + x + 5 + 7 + y + 3 + 1 = 25 \Rightarrow x + y = 8$ .
Mean condition: $\text{Mean} = \frac{\Sigma fx}{\Sigma f} = 35 \Rightarrow 605 + 15x + 45y = 875 \Rightarrow (\div 15)\ x + 3y = 18$ . Solving the pair gives $x = 3, y = 5$ .

Common Mistakes

Using class boundaries (0, 10, 20, …) instead of midpoints when computing Σfx.
Arithmetic slips while summing the constant products (should total 605).
Not simplifying 15x + 45y = 270 to x + 3y = 18 before elimination, making the algebra heavier.

Real-World Relevance

This back-calculation technique — recovering missing frequencies from a reported mean and total — is exactly how incomplete survey or census tables are reconstructed from published summary statistics.

Common Mistakes

1Using class boundaries (0, 10, 20 …) instead of class marks/midpoints (5, 15, 25 …) when computing $\Sigma fx$ , which makes every subsequent value wrong.
2Arithmetic error in adding the known products ( $5 + 125 + 245 + 165 + 65 = 605$ ); an incorrect constant propagates into both equations.
3Setting up only one equation and trying to guess the second frequency, instead of using both the total-frequency and mean conditions to form two simultaneous equations.

Interesting Facts

This 'reverse' use of the mean — knowing the average and total to recover missing data — is exactly how statisticians and auditors reconstruct suppressed or missing cells in published data tables that only report summary figures.

The direct method used here works because the class mark is the best single representative of every value inside an interval; it assumes values are evenly spread around the midpoint, which is why the answer is an estimate of the true mean.

For grouped data, the direct method, the assumed-mean method, and the step-deviation method always give the same mean — students can switch to the step-deviation method to reduce large-number arithmetic when class marks are big.

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Frequently Asked Questions

Why do we use class marks instead of the class limits?

In grouped data every value inside an interval is represented by a single number — the class mark, which is the midpoint $\frac{\text{lower limit} + \text{upper limit}}{2}$ . For 0-10 the class mark is 5, for 10-20 it is 15, and so on. Using the class limits (0, 10, 20) instead of the midpoints is a serious error that makes the mean calculation incorrect.

Why do we need two equations here?

There are two unknown frequencies, x and y, so we need two independent equations to solve for them. The total-frequency condition $\Sigma f = 25$ gives $x + y = 8$ , and the mean condition $\text{Mean} = \frac{\Sigma fx}{\Sigma f} = 35$ gives $x + 3y = 18$ . Solving these two simultaneous equations gives the unique values $x = 3$ and $y = 5$ .

Could I solve this with the step-deviation method instead?

Yes. The step-deviation method gives exactly the same answer and can reduce arithmetic when class marks are large. However, since two frequencies are unknown, you would still need both the total-frequency equation and the mean equation; the method only changes how the mean equation is computed, not the overall two-equation approach.