Statistics (Mean, Median, Mode) Questions and Answers

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.

Answer: B

Quartiles are values that divide a data set into four equal parts.

• Q1 (First Quartile): The 25th percentile.
• Q2 (Second Quartile): The 50th percentile, which is the definition of the median.
• Q3 (Third Quartile): The 75th percentile.

Answer: C

Current total age of 5 members = 5 × 21 = 105 years.

5 years ago, the sum of their ages would have been 105 - (5 members × 5 years) = 105 - 25 = 80 years.

At the birth of the youngest member (5 years ago), there were only 4 members in the family (the youngest was just born, age 0).

Average age at that time = Total sum of ages of the 4 members / 4. This is tricky. The sum of ages of the 4 older members 5 years ago was 80. The average age was 80/4 = 20 years. Let's re-verify. Let ages be a,b,c,d,e. (a+b+c+d+e)/5 = 21. Sum = 105. Youngest e=5. At birth of e, ages were a-5, b-5, c-5, d-5. The sum of their ages was (a+b+c+d)-20. We know a+b+c+d = 105-5=100. So sum was 100-20=80. Average = 80/4 = 20.

Answer: A

Correct answer: 9.000

For a frequency distribution the mean is the total of \(f_i x_i\) divided by the total frequency.

\[\sum f_i x_i = (4\times 5) + (6\times 10) + (9\times 10) + (10\times 7) + (15\times 8)\]

\[= 20 + 60 + 90 + 70 + 120 = 360\]

\[\sum f_i = 5 + 10 + 10 + 7 + 8 = 40\]

\[\text{Mean} = \frac{360}{40} = 9\]

Averaging the values alone, ignoring their frequencies, would give \(\tfrac{44}{5} = 8.8\) - the usual error here.

Answer: A

The mean of a frequency distribution is calculated using the formula: Mean = \(\frac{\sum f_i x_i}{\sum f_i}\).

First, we calculate the product of each value and its frequency (fx):

• 4 × 5 = 20
• 6 × 10 = 60
• 9 × 10 = 90
• 10 × 7 = 70
• 15 × 8 = 120

Next, we sum these products: \(\sum f_i x_i = 20 + 60 + 90 + 70 + 120 = 360\).

Then, we sum the frequencies: \(\sum f_i = 5 + 10 + 10 + 7 + 8 = 40\).

Finally, we calculate the mean: Mean = \(\frac{360}{40} = 9\).

Answer: C

The mode is, by definition, the most frequent value *in the data set*. Therefore, it must be one of the numbers present in the set.

The mean and median can be values that do not actually appear in the data set. For example, for the set {2, 4}, the mean is 3 and the median is 3, neither of which is in the set.

Answer: B

This is a weighted average problem.

Total marks for section A = 32 × 60 = 1920.

Total marks for section B = 40 × 33 = 1320.

Combined total marks = 1920 + 1320 = 3240.

Combined total students = 32 + 40 = 72.

Combined average = \(\frac{3240}{72} = 45\).

Answer: B

A fundamental property of the arithmetic mean is that the sum of the deviations of the data points from their mean is always zero.

We can verify: (1-4) + (3-4) + (4-4) + (5-4) + (7-4) + (4-4)

= (-3) + (-1) + 0 + 1 + 3 + 0 = 0.

Answer: D

The median of 9 observations is the 5th term when arranged in order. Its value is 20.5.

The change is made to the largest 4 observations, which are the 6th, 7th, 8th, and 9th terms.

Since these changes do not affect the value or position of the 5th term, the median of the new set will remain the same.

Answer: C

Correct answer: 27.00

The modal class is the one with the greatest frequency: 20–30, with a frequency of 15.

\[\text{Mode} = L + \frac{f_m - f_1}{2f_m - f_1 - f_2} \times h\]

Here \(L = 20\), \(h = 10\), \(f_m = 15\), the preceding frequency \(f_1 = 8\) and the succeeding frequency \(f_2 = 12\).

\[\text{Mode} = 20 + \frac{15 - 8}{2(15) - 8 - 12} \times 10 = 20 + \frac{7}{10} \times 10 = 27\]

The mode sits above the midpoint of its class because the class after it is heavier than the class before it.