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 standard deviation of 10 observations ${x_{1\,\,}}{x_2}\,,\,....\,\,{x_{10}}$ is $4$ that of another set of $10$ observations ${y_{1\,\,}}{y_2}\,,\,....\,\,{y_{10}}$ is $3$ and also ${X_i}\, = \,\left( {{x_i}\, - \,\overline x } \right)\,\left( {{y_i}\, - \,\overline y } \right)\,,\,\overrightarrow x $ is mean of all $x_{ i }^{ 's }\ $ and $y$ is mean of all $y_{ i }^{ 's }\,\sum\limits_{i = 1}^{10} {{X_i}} \, = \,80 $ then standard deviation of ten observation $\left( {{x_1}\, - \,{y_1}} \right)\,,\,\left( {{x_2}\, - \,{y_2}} \right)\,,.......\left( {{x_{10}}\, - {y_{10}}} \right)$ is
  1. $1$
  2. $3$
  3. $5$
  4. $\sqrt 5 $
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice

Professor Buckingham, a chemist and teacher at a community college, is organising his graduated cylinders in the hopes of keeping his office tidy and setting a good example for his students. He has beakers with diameters, in inches, of $\displaystyle\frac{1}{2}, \frac{3}{4}, \frac{4}{5}, 1$ and $\displaystyle\frac{5}{4}$. With his original five cylinders, professor Buckingham realises that he is missing a cylinder necessary for his upcoming lab demonstration for Thursday's class. He remembers that the cylinder he needs when added to the original five, will create a median diameter value of $\displaystyle\frac{9}{10}$ for the set of six total cylinders. He also knows that the measure of the sixth cylinder will exceed the value of the range of the current five cylinders by a width of anywhere from $\displaystyle\frac{1}{4}$ inches to $\displaystyle\frac{1}{2}$ inches, inclusive. Based on the data, what is one possible value of y, the diameter of this missing sixth cylinder?
  1. $1 \leq y\leq 1.25$
  2. $2 \leq y\leq 2.25$
  3. $3 \leq y\leq 3.25$
  4. $4 \leq y\leq 4.25$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The current range is 1.25-0.5=0.75 inches. The sixth diameter must therefore be between 1 and 1.25 inches, and y=1 also gives a median of 0.9 for the six values.

Multiple choice

In TVS-2 batch every student was weighed on two weighing machines A and B Machine A and machine B both have some technical errors Machine A gives the reading equal to four times the original weight of a student and machine B gives the reading equal to 5 kg more than the actual weight If variances calculated by readings of machines A and B were $\displaystyle V_{A}: : : and: : : V_{B} $ respectively and mean deviations are $\displaystyle M_{A}: : : and: : : M_{B} $ respectively then
  1. $\displaystyle M_{A}=4M_{B}$
  2. $\displaystyle \frac{V_{A}}{V_{B}}=16\left ( \frac{M_{A}}{M_{B}} \right )$
  3. $\displaystyle \frac{V_{A}}{V_{B}}= \frac{M_{A}}{M_{B}}$
  4. $\displaystyle V_{A}=4V_{B}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice

If all the class marks are reduced by 5 and divided by 10 and then the mean of the numbers so obtained is found out to be "x". How is "x" related to actual mean "m"?

  1. x = m

  2. x = 10 m - 5

  3. 10x = m - 5

  4. x = 50 m - 10
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the original values be xi. The new values are yi = (xi - 5) / 10. The mean x = (1/n) * sum(yi) = (1/n) * sum((xi - 5) / 10) = (1/10) * ((1/n) * sum(xi) - 5) = (m - 5) / 10. Therefore, 10x = m - 5.

Multiple choice

Given table shows the marks scored by various students in a class test. Find the mean of the given data. Marks 41 - 50 51 - 60 61 - 70 71 - 80 81 - 90 91 - 100 Frequency 8 17 12 14 20 3

  1. 69.55

  2. 72.48

  3. 88.67

  4. 97.24
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Use the midpoints of the marks: 45.5, 55.5, 65.5, 75.5, 85.5, 95.5. Multiply by frequencies: 8*45.5=364, 17*55.5=943.5, 12*65.5=786, 14*75.5=1057, 20*85.5=1710, 3*95.5=286.5. Sum = 5147. Total frequency = 74. Mean = 5147/74 = 69.554.

Multiple choice

9.4, 9.9, 9.9, 9.9, 10.0, 10.2, 10.2, 10.5 The mean and the standard deviation of the 8 numbers shown are 10 and 0.3, respectively. What percentage of the 8 number's are within 1 standard deviation?

  1. 90%

  2. 85%

  3. 80%

  4. 75%

  5. 70%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

One standard deviation from the mean gives the interval 10 - 0.3 to 10 + 0.3, or 9.7 to 10.3. Six values lie in this interval, while 9.4 and 10.5 do not. Thus, the percentage is 6/8 x 100 = 75%.

Multiple choice

Find the values of 'x' and 'y' in the following data. Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 Frequency 13 17 x 15 y 20 The mean of the data is 31.5 and the number of observations is 80.

  1. x = 7, y = 8

  2. x = 3, y = 12

  3. x = 4, y = 11

  4. x = 5, y = 10
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The sum of frequencies is 13 + 17 + x + 15 + y + 20 = 80, so x + y = 15. Using the mean formula with class midpoints (5, 15, 25, 35, 45, 55), the equation 31.5 * 80 = 65 + 255 + 25x + 525 + 45y + 1100 leads to 25x + 45y = 570, or 5x + 9y = 114. Solving the system yields x = 5 and y = 10.

Multiple choice

Find the mean and median of the following data: (Mean by step deviation method) Age of children 5 - 15 15 - 25 25 - 35 35 - 45 45 - 55 55 - 65 Number of children 5 10 20 30 15 20

  1. 40 and 42

  2. 41 and 40

  3. 40 and 40

  4. 38 and 40
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total frequency = 100. Mean = sum(f*x)/sum(f). Midpoints: 10, 20, 30, 40, 50, 60. f*x: 50, 200, 600, 1200, 750, 1200. Sum = 4000. Mean = 4000/100 = 40. Median class is where cumulative frequency reaches 50, which is 35-45. Median = L + ((N/2 - cf)/f) * h = 35 + ((50 - 35)/30) * 10 = 35 + (15/30)*10 = 35 + 5 = 40.