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 frequency distribution of marks obtained by $28$ students in a test carrying $40$ marks is given below Marks 0-10 10-20 20-30 30-40 Number of students 6 $x$ $y$ 6 If the mean of the above data is $20$, then the difference between $x$ and $y$ is
  1. $3$
  2. $2$
  3. $1$
  4. $0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Midpoints: 5, 15, 25, 35. Mean = (5*6 + 15*x + 25*y + 35*6) / 28 = 20. 30 + 15x + 25y + 210 = 560. 15x + 25y = 320. Divide by 5: 3x + 5y = 64. Also 6+x+y+6 = 28, so x+y = 16. Solve: 3x + 5(16-x) = 64 -> 3x + 80 - 5x = 64 -> -2x = -16 -> x=8. Then y=8. Difference x-y = 0.

Multiple choice

The numbers $3,5,7,4 $ have frequencies $x,x+4,x-3,x+8.$ If their arithmetic mean is $4 $, the value of $x$ is
  1. $\displaystyle \frac{5}{3}$
  2. $\displaystyle \frac{7}{4}$
  3. $\displaystyle \frac{2}{3}$

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

Mean = (Sum of (value * frequency)) / (Sum of frequencies) = 4. (3x + 5(x+4) + 7(x-3) + 4(x+8)) / (x + x+4 + x-3 + x+8) = 4. (3x + 5x + 20 + 7x - 21 + 4x + 32) / (4x + 9) = 4. (19x + 31) / (4x + 9) = 4. 19x + 31 = 16x + 36. 3x = 5. x = 5/3.

Multiple choice

If the mean of a set of observations $\displaystyle x_{1},x_{2},......,x_{10}$ is 20, then the mean of $\displaystyle x_{1}+4,x_{2}+8,x_{3}+12,....,x_{10}+40$ is
  1. $34$
  2. $42$
  3. $38$
  4. $40$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Original mean = (Sum of x_i) / 10 = 20, so Sum = 200. New sum = (x_1+4) + (x_2+8) + ... + (x_10+40) = Sum(x_i) + (4+8+...+40). Sum of arithmetic series = (10/2) * (4+40) = 5 * 44 = 220. New sum = 200 + 220 = 420. New mean = 420 / 10 = 42.

Multiple choice

A data set of $\displaystyle n$ observations has mean $\displaystyle 2\overset { - }{ x } $ while another data set of $\displaystyle 2n$ observations has mean $\displaystyle \overset { - }{ x } $. The mean of the combined data set of $\displaystyle 3n$ observations will be equal to:
  1. $\displaystyle \overset { - }{ x } $
  2. $\displaystyle \frac { 3 }{ 2 } \overset { - }{ x } $
  3. $\displaystyle \frac { 2 }{ 3 } \overset { - }{ x } $
  4. $\displaystyle \frac { 4 }{ 3 } \overset { - }{ x } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Sum of first set = n * 2x_bar. Sum of second set = 2n * x_bar. Total sum = 2n*x_bar + 2n*x_bar = 4n*x_bar. Total observations = 3n. Combined mean = 4n*x_bar / 3n = 4/3 * x_bar.

Multiple choice

The mean of twenty observations $\displaystyle { x }{ 1 },{ x }{ 2 },{ x }{ 3 },{ x }{ 4 },...{ x }_{ 20 }$ is $\displaystyle m$. If each of the first ten observations is increased by 8 and each of the next 10 observations is decreased by 8, then the new mean is $\displaystyle n$. The difference $\displaystyle m - n$ equals
  1. $16$
  2. $20$
  3. $0$
  4. $8$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Initial sum = 20 * m. New sum = (sum of first 10 + 80) + (sum of next 10 - 80) = original sum. New mean n = original sum / 20 = m. So m - n = 0.