A distribution where the Mean = 60, Median = 65, and Mode = 70 is:
Answer: C
We compare the values of the central tendencies.
Here, Mean (60) < Median (65) < Mode (70).
When the mean is pulled to the left of the median and mode, the distribution has a long tail on the left side and is said to be negatively skewed.
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What is the mode of the first 10 prime numbers?
Answer: D
The first ten primes are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29 - all different.
With every value occurring exactly once there is no mode.
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The average of 5 quantities is 6. The average of 3 of them is 4. What is the average of the remaining two quantities?
Answer: C
Total sum of the 5 quantities = 5 × 6 = 30.
Sum of 3 of these quantities = 3 × 4 = 12.
Sum of the remaining two quantities = 30 - 12 = 18.
Average of the remaining two quantities = \(\frac{18}{2} = 9\).
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If each observation of a data set is multiplied by 3, the mean of the new data set is:
Answer: B
The mean is a linear function of the data, so multiplying every observation by 3 multiplies the mean by 3.
\[ \frac{\sum 3x_{i}}{n} = 3 \times \frac{\sum x_{i}}{n} \]
The variance, by contrast, would be multiplied by 9.
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A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:
Answer: D
If a month of 30 days begins with a Sunday, it will have 5 Sundays and 25 other days.
Total visitors on Sundays = 5 × 510 = 2550.
Total visitors on other days = 25 × 240 = 6000.
Total visitors in the month = 2550 + 6000 = 8550.
Average visitors per day = \(\frac{8550}{30} = 285\).
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Find the median of the following frequency distribution:
| Value (x) | 10 | 12 | 14 | 16 | 18 |
|---|---|---|---|---|---|
| Frequency (f) | 3 | 5 | 6 | 4 | 2 |
Answer: C
First, find the total frequency (N) and the cumulative frequency (cf).
N = 3+5+6+4+2 = 20.
The median term is the average of the N/2 and (N/2)+1 terms, i.e., the 10th and 11th terms.
Both the 10th and 11th terms fall in the group where the value is 14. Therefore, the median is 14.
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The sum of the deviations of a set of n values from their arithmetic mean is:
Answer: A
\[ \sum (x_{i} - \bar{x}) = \sum x_{i} - n\bar{x} = n\bar{x} - n\bar{x} = 0 \]
The deviations always cancel, which is exactly why the mean is the balance point of the data - and why absolute or squared deviations are used to measure spread.
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The average marks of a student in 5 subjects is 88. He scored 82, 85, 90, and 92 in four subjects. What are his marks in the fifth subject?
Answer: D
Total marks in 5 subjects = Average × Number of subjects = 88 × 5 = 440.
Sum of marks in four subjects = 82 + 85 + 90 + 92 = 349.
Marks in the fifth subject = Total marks - Sum of four subjects = 440 - 349 = 91.
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If the mean of a, b, c is M and ab + bc + ca = 0, then the mean of a², b², c² is:
Answer: B
Given \(\frac{a+b+c}{3} = M \Rightarrow a+b+c=3M\).
We know the algebraic identity: \((a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)\).
Substitute the given values:
\((3M)^2 = (a^2+b^2+c^2) + 2(0)\)
\(9M^2 = a^2+b^2+c^2\).
The mean of a², b², c² is \(\frac{a^2+b^2+c^2}{3} = \frac{9M^2}{3} = 3M^2\).
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A car travels from A to B at a speed of 40 km/hr and returns from B to A at a speed of 60 km/hr. What is the average speed for the entire journey?
Answer: A
When the distance is constant for two speeds x and y, the average speed is the Harmonic Mean of the speeds, not the Arithmetic Mean.
Average Speed = \(\frac{2xy}{x+y}\).
Average Speed = \(\frac{2 \times 40 \times 60}{40+60} = \frac{4800}{100} = 48\) km/hr.
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