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

If the mean of the observations: x,x+3,x+5,x+7,x+10 is 9 the mean of the last three observations is
  1. $\displaystyle 12\frac{1}{3}$
  2. $\displaystyle 12\frac{2}{3}$
  3. $\displaystyle 11\frac{1}{3}$
  4. $\displaystyle 13\frac{2}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The mean of x, x+3, x+5, x+7, x+10 is (5x + 25) / 5 = x + 5. Given x + 5 = 9, we find x = 4. The last three observations are x+5, x+7, x+10, which are 9, 11, 14. Their mean is (9 + 11 + 14) / 3 = 34 / 3 = 11 1/3.

Multiple choice

Marks 10-20 20-30 30-40 40-50 50-60 Frequency 4 10 20 40 50 The following frequency distribution showing the marks obtained by $124$ students in Economy at a certain school. Find the arithmetic mean using direct method.
  1. $50$ marks
  2. $80$ marks
  3. $30$ marks
  4. $45$ marks
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Midpoints (x): 15, 25, 35, 45, 55. Frequencies (f): 4, 10, 20, 40, 50. Sum(f) = 124. Sum(fx) = (4*15) + (10*25) + (20*35) + (40*45) + (50*55) = 60 + 250 + 700 + 1800 + 2750 = 5560. Mean = 5560 / 124 = 44.83, which rounds to 45.

Multiple choice

Class 0-20 20-40 40-60 60-80 80-100 Frequency 19 $f_1$ 32 $f_2$ 19 Total 120 Mean of the above frequency table is given as $50$. Find $f_1$ and $f_2$.
  1. $28$ and $22$
  2. $25$ and $25$
  3. $24$ and $26$
  4. $20$ and $30$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total frequency = 19 + f1 + 32 + f2 + 19 = 120, so f1 + f2 = 50. Mean = sum(fi*xi)/120 = 50. Midpoints are 10, 30, 50, 70, 90. (19*10 + f1*30 + 32*50 + f2*70 + 19*90) / 120 = 50. 190 + 30f1 + 1600 + 70f2 + 1710 = 6000. 30f1 + 70f2 = 2500, or 3f1 + 7f2 = 250. Substituting f1 = 50 - f2: 3(50-f2) + 7f2 = 250 => 150 + 4f2 = 250 => 4f2 = 100 => f2 = 25. Then f1 = 25.

Multiple choice

The following table gives the marks obtained by $300$ students of a management course. Find the median of the distribution. Marks obtained No. of Students $0-15$ $26$ $15-30$ $34$ $30-45$ $64$ $45-60$ $76$ $60-75$ $60$ $75-90$ $30$ $90-100$ $10$
  1. $62.96$ marks
  2. $62.13$ marks
  3. $50.13$ marks
  4. $50.96$ marks
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total students = 300. Median class is where cumulative frequency reaches 150. CF: 26, 60, 124, 200. Median class is 45-60. Median = L + ((N/2 - CF)/f)*h = 45 + ((150 - 124)/76)*15 = 45 + (26/76)*15 = 45 + 5.13 = 50.13.