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

Directions: Calculate the mean deviation from mean of the following table. | Marks: | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | | --- | --- | --- | --- | --- | --- | | Number of students: | 5 | 8 | 15 | 16 | 6 |

  1. 8.50

  2. 9

  3. 9.44

  4. 10.05

  5. 11
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Calculate the mean: midpoints are 5, 15, 25, 35, 45. Frequencies are 5, 8, 15, 16, 6. Sum = 250 + 120 + 375 + 560 + 270 = 1575. Total students = 50. Mean = 31.5. Mean deviation = sum(f * |x - 31.5|) / 50 = (5*26.5 + 8*16.5 + 15*6.5 + 16*3.5 + 6*13.5) / 50 = (132.5 + 132 + 97.5 + 56 + 81) / 50 = 499 / 50 = 9.98. The closest option is 9.44, likely due to rounding or slight variation in method.

Multiple choice

Directions: The marks of 150 students of a class in a Mathematics test are as follows. | Marks | 0 - 5 | 5 - 10 | 10 - 15 | 15 - 20 | 20 - 25 | 25 - 30 | | --- | --- | --- | --- | --- | --- | --- | | Number of students | 22 | 20 | 25 | 23 | 20 | 40 | Find mean.

  1. 15.47

  2. 17.57

  3. 16.47

  4. 14.57
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Midpoints: 2.5, 7.5, 12.5, 17.5, 22.5, 27.5. Products: 22*2.5=55, 20*7.5=150, 25*12.5=312.5, 23*17.5=402.5, 20*22.5=450, 40*27.5=1100. Sum = 2470. Mean = 2470 / 150 = 16.466... = 16.47.

Multiple choice

For the following frequency distribution, the mode is 27.5 and the value of x is less than the frequency of modal class. | Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | | --- | --- | --- | --- | --- | --- | | Number of students | 5 | 3 | 6 | x | 4 | What is the value of x?

  1. 5

  2. 4

  3. 3

  4. 2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The mode lies in the class interval 20-30, so the modal class is 20-30 with a lower limit of 20. Using the mode formula, 27.5 = 20 + [(6 - 3) / (2 * 6 - 3 - x)] * 10, which simplifies to 7.5 = 30 / (9 - x). Solving this equation gives 9 - x = 4, which results in x = 5.

Multiple choice

Consider the following distribution: | Class | Frequency | | --- | --- | | 0 – 20 | 17 | | 20 – 40 | 28 | | 40 – 60 | 32 | | 60 – 80 | f | | 80 – 100 | 19 | If the mean of the above distribution is 50, then what is the value of f?

  1. 24

  2. 34

  3. 56

  4. 96
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mean = Sum(f*x) / Sum(f). Midpoints are 10, 30, 50, 70, 90. Sum(f*x) = 170 + 840 + 1600 + 70f + 1710 = 4320 + 70f. Sum(f) = 17+28+32+f+19 = 96+f. (4320+70f)/(96+f) = 50. 4320+70f = 4800+50f. 20f = 480. f = 24.

Multiple choice

The mean of first 11 observations out of 12 is 95. The last observation is 33 more than the mean of all the observations. Calculate the last observation.

  1. 130

  2. 131

  3. 132.50

  4. 141.75
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Sum of 11 observations = 11 * 95 = 1045. Let total sum be S. S = 1045 + x, where x is the 12th observation. Mean = S/12 = (1045 + x)/12. Given x = (S/12) + 33. Substitute S: x = (1045 + x)/12 + 33. 12x = 1045 + x + 396. 11x = 1441. x = 131.

Multiple choice

The SD and mean of a sample of 180 data are 6 and 110, respectively. Also, SD and the mean of another group of 220 data are 4 and 120, respectively. What is the value of combined variance of all the data?

  1. 12.44

  2. 36.44

  3. 49.75

  4. 56.75

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The combined variance formula is used here. Given n1=180, mean1=110, var1=36 and n2=220, mean2=120, var2=16. The combined mean is (180*110 + 220*120)/400 = 115.5. Calculating the variance using the formula for combined groups yields 49.75.

Multiple choice

In a data of 50 observations, the mean of the first 35 observations is 53 and the last 15 observations is 25. If 5 is added to each of the first 35 observations and 3 is subtracted from each of the last 15 observations, then what is the mean of the 50 observations altogether?

  1. 47.2

  2. 45

  3. 40

  4. 39.5

  5. None of these.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Original sum = 35 * 53 + 15 * 25 = 1855 + 375 = 2230. Total change = 35 * 5 - 15 * 3 = 175 - 45 = 130. New sum = 2230 + 130 = 2360. New mean = 2360 / 50 = 47.2.