Quantitative Aptitude

Statistics

606 Questions
All Ungrouped data dispersion and consistency 209 Unknown terms from given average 124 Grouped data central tendencies 101 Merging means and variances of groups 53 Replacement change in average 27 Alligation and weighted average 21 Data interpretation from tables and charts 20 Net successive percentage change 19 Grouped data variance via step deviation 18 Combination of multiple ratios 1

Statistics Questions

Multiple choice

The mean of two samples of sizes $200 $ and $300$ were found to be $25$ and $10$ respectively. Their standard deviations were $ 3 $ and $4$ respectively. The varience of combined sample size of $500$ is
  1. $64$
  2. $65.2$
  3. $67.2$
  4. $64.2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The combined variance is calculated using the formula for combined mean and variance. First, find the combined mean: (200*25 + 300*10) / 500 = 16. Then, use the formula: Variance = (n1(s1^2 + d1^2) + n2(s2^2 + d2^2)) / (n1 + n2), where d1 = 25-16 = 9 and d2 = 10-16 = -6. This yields (200*(9+81) + 300*(16+36)) / 500 = (18000 + 15600) / 500 = 33600 / 500 = 67.2.

Multiple choice

If the sum of the mode and mean of the certain frequency distribution in $129$ and the median of the observations is $63$, mode and mean are respectively
  1. $69$ and $60$
  2. $65$ and $64$
  3. $68$ and $61$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Empirical formula: Mode = 3*Median - 2*Mean. Given Mode + Mean = 129 and Median = 63. Mode = 3(63) - 2*Mean = 189 - 2*Mean. Substitute into Mode + Mean = 129: (189 - 2*Mean) + Mean = 129 => 189 - Mean = 129 => Mean = 60. Mode = 129 - 60 = 69.

Multiple choice

The mean square deviation of set of $n$ observations ${ x }{ 1 },{ x }{ 2 },.....{ x }{ n }$ about a point $c$ is defined as $\displaystyle \frac { 1 }{ n } \sum _{ i=1 }^{ n }{ { \left( { x }{ i }-c \right) }^{ 2 } } $ The mean square deviation about $-2$ and $2$ are $18$ and $10$ respectively, then standard deviation of this set of observations is
  1. $3$
  2. $2$
  3. $1$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mean square deviation about c is E[(X-c)^2] = E[X^2] - 2cE[X] + c^2. Given E[(X+2)^2] = 18 and E[(X-2)^2] = 10. Expanding: E[X^2] + 4E[X] + 4 = 18 and E[X^2] - 4E[X] + 4 = 10. Subtracting: 8E[X] = 8, so E[X] = 1. Then E[X^2] + 4 + 4 = 18, so E[X^2] = 10. Variance = E[X^2] - (E[X])^2 = 10 - 1 = 9. Standard deviation = sqrt(9) = 3.

Multiple choice

In a set of $2n$ distinct observation, each of the observation below the median of all the observations is increased by $5$ and each of the remaining observations is decreased by $3$. Then the mean of the new set of observations :
  1. Increases by $1$.
  2. Decreases by $1$.
  3. Decreases by $2$.
  4. Increases by $2$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In 2n observations, n are below the median and n are above/at the median. Sum change = n*5 + n*(-3) = 2n. New mean change = 2n / 2n = 1.

Multiple choice

Mean of 'n' items is $\bar{x}$. If these n items are successively increased by $2, 2^2, 2^3$,.... $2^n$, then the new mean is?
  1. $\bar{x}+\dfrac{2^{n+1}}{n}$
  2. $\bar{x}+\dfrac{2^{n+1}}{n}-\dfrac{2}{n}$
  3. $\bar{x}+\dfrac{2^n}{n}$
  4. $\bar{x}+2^n$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The original sum of n items is n * x_bar. The new sum is (n * x_bar) + (2 + 2^2 + ... + 2^n). The sum of the geometric series 2 + 2^2 + ... + 2^n is 2 * (2^n - 1) / (2 - 1) = 2^(n+1) - 2. The new mean is (n * x_bar + 2^(n+1) - 2) / n = x_bar + (2^(n+1) - 2) / n, which simplifies to x_bar + 2^(n+1)/n - 2/n.