JEE Main 2023 · 13 April · morning · Numerical answer · +4 / −0
Let the mean of the data
be 5. If m and σ² are respectively the mean deviation about the mean and the variance of the data, then (3 α)/(m+σ²) is equal to __________
| x | 1 | 3 | 5 | 7 | 9 |
|---|---|---|---|---|---|
| Frequency (f) | 4 | 24 | 28 | α | 8 |
Answer and solution
Answer 8
Solution
Maths sir · 3 steps
Use the mean to find α first, then work with deviations from 5 to get m and σ² together.
- Find alpha. Total frequency is N = 64 + α and Σ fx = 4 + 72 + 140 + 7α + 72 = 288 + 7α. Setting the mean to 5 gives 288 + 7α = 320 + 5α, so α = 16 and N = 80.
x f |x - 5| f |x - 5| f (x - 5)² 1 4 4 16 64 3 24 2 48 96 5 28 0 0 0 7 16 2 32 64 9 8 4 32 128 Total 80 not needed 128 352 - Mean deviation and variance. m = 128/80 = 1.6 and σ² = 352/80 = 4.4. Because 5 is the mean, the variance is the average of the squared deviations from 5.
3α/(m + σ²) = (3 × 16)/(1.6 + 4.4) = 48/6
- So. The required value is 8.
The idea to keep. Once the mean is fixed, one column of deviations x - x̄ gives you both measures: average the absolute values for mean deviation and average the squares for variance.
Answer from the NTA final answer key. Solution by PrepWiser.