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 deviation from the data $3, 10, 10, 4, 7, 10, 5$.
  1. $3$
  2. $2$
  3. $3.75$
  4. $2.57$

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

First, find the mean: (3+10+10+4+7+10+5) / 7 = 49 / 7 = 7. Then, calculate the absolute deviations from the mean: |3-7|=4, |10-7|=3, |10-7|=3, |4-7|=3, |7-7|=0, |10-7|=3, |5-7|=2. The mean of these deviations is (4+3+3+3+0+3+2) / 7 = 18 / 7 = 2.57.

Multiple choice

If mean of the following frequency distribution is $1.46$ then find the unknown frequency. x: 0 1 2 3 4 5 Total f: 46 ... ... 25 10 5 200
  1. $38$ and $76$
  2. $35$ and $72$
  3. $38$ and $71$
  4. $32$ and $72$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum of the frequencies is 200, which gives the equation f1 + f2 = 114. Using the formula for the mean, we get another equation f1 + 2*f2 = 152. Solving this system of linear equations yields f1 = 76 and f2 = 38.

Multiple choice

Let $\overline x$, M and $\sigma^2$ be respectively the mean mode and variance of n observations $x_1, x_2, ....., x_n$ and $d_i=-x_1-a, i=1, 2, ....., n,$ where a is any number. Statement I : Variance of $d_1, d_1, ....., d_n$ is $\sigma^2$ Statement II : Mean and mode of $d_1, d_2, ....., d_n$ are $-\overline x-a$ and $-M-a$, respectively

  1. Statement I and Statement II are both false

  2. Statement I and Statement II are both true

  3. Statement I is true and Statement II is false

  4. Statement I is false and Statement II is true
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice

A random variable $X$ takes the values $0,\ 1,\ 2,\ 3$ and its mean is$1.3$. If $P(X=3)=2P(X=1)$ and $P(X=2)=0.3$ , then $P(X=0)=$
  1. $0.1$
  2. $0.2$
  3. $0.3$
  4. $0.4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Sum of probabilities = 1. P(0) + P(1) + P(2) + P(3) = 1. P(2) = 0.3, so P(0) + P(1) + P(3) = 0.7. P(3) = 2P(1). So P(0) + 3P(1) = 0.7. Mean = 0*P(0) + 1*P(1) + 2*P(2) + 3*P(3) = 1.3. P(1) + 2(0.3) + 3(2P(1)) = 1.3. P(1) + 0.6 + 6P(1) = 1.3. 7P(1) = 0.7. P(1) = 0.1. Then P(0) + 3(0.1) = 0.7, so P(0) = 0.4.

Multiple choice

The mean of $5$ observation is $5$ and their variance is $12.4$. If three of the observations are $1, 2$ and $6$; then the mean deviation from the mean of the data is:
  1. $2.5$
  2. $2.6$
  3. $2.8$
  4. $2.4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Mean = 5, n=5. Sum = 25. Observations: 1, 2, 6, x, y. 1+2+6+x+y = 25 => x+y = 16. Variance = 12.4 = [Sum(x_i^2)/n] - mean^2. 12.4 = [1^2 + 2^2 + 6^2 + x^2 + y^2]/5 - 25. 37.4 * 5 = 187 = 41 + x^2 + y^2. x^2 + y^2 = 146. Solving x+y=16 and x^2+y^2=146: (x+y)^2 = 256 = x^2+y^2+2xy => 256 = 146 + 2xy => 2xy = 110 => xy=55. Roots are 5 and 11. Data: 1, 2, 6, 5, 11. Mean deviation = (|1-5| + |2-5| + |6-5| + |5-5| + |11-5|) / 5 = (4+3+1+0+6)/5 = 14/5 = 2.8.