Q: For a given data with mean 60 and median 50, what is the mode?

Using the formula $\text{Mode} = 3 \text{ Median} - 2 \text{ Mean}$, we substitute $\text{Median} = 50$ and $\text{Mean} = 60$. $\text{Mode} = 3(50) - 2(60) = 150 - 120 = 30$. Thus, option A is correct.

Q: The lower limit of the median class of the following data is:\nClass: 0-10, 10-20, 20-30, 30-40, 40-50\nFrequency: 3, 9, 15, 30, 18

Total frequency $N = 3 + 9 + 15 + 30 + 18 = 75$. $N/2 = 37.5$. Cumulative frequencies are: 3, 12, 27, 57, 75. The cumulative frequency just greater than 37.5 is 57, which corresponds to the class 30-40. The lower limit is 30. Therefore, option C is correct.

Q: If the mean of observations $x, x+3, x+5, x+7,$ and $x+10$ is 9, then the mean of the last three observations is:

Sum of observations = $x + (x+3) + (x+5) + (x+7) + (x+10) = 5x + 25$. Mean = $(5x + 25)/5 = x + 5 = 9 \implies x = 4$. Last three observations are $x+5, x+7, x+10$, which are $9, 11, 14$. Their mean = $(9+11+14)/3 = 34/3 = 11.33$. Wait, let's recalculate: $x+5=9, x+7=11, x+10=14$. Sum =

Question 1

What is the empirical relationship between mean, median, and mode?

Accepted Answer

The empirical relationship between mean, median, and mode is given by the formula: $3 \text{ Median} = \text{Mode} + 2 \text{ Mean}$, which can also be written as $\text{Mode} = 3 \text{ Median} - 2 \text{ Mean}$. Therefore, option C is correct.

Question 2

Which of the following cannot be determined graphically?

Accepted Answer

The mean cannot be determined graphically. Median can be found using ogives, and mode can be found using a histogram. Therefore, option B is correct.

Question 3

For a given data with mean 60 and median 50, what is the mode?

Accepted Answer

Using the formula $\text{Mode} = 3 \text{ Median} - 2 \text{ Mean}$, we substitute $\text{Median} = 50$ and $\text{Mean} = 60$. $\text{Mode} = 3(50) - 2(60) = 150 - 120 = 30$. Thus, option A is correct.

Question 4

The lower limit of the median class of the following data is:
Class: 0-10, 10-20, 20-30, 30-40, 40-50
Frequency: 3, 9, 15, 30, 18

Accepted Answer

Total frequency $N = 3 + 9 + 15 + 30 + 18 = 75$. $N/2 = 37.5$. Cumulative frequencies are: 3, 12, 27, 57, 75. The cumulative frequency just greater than 37.5 is 57, which corresponds to the class 30-40. The lower limit is 30. Therefore, option C is correct.

Question 5

If the mean of observations $x, x+3, x+5, x+7,$ and $x+10$ is 9, then the mean of the last three observations is:

Accepted Answer

Sum of observations = $x + (x+3) + (x+5) + (x+7) + (x+10) = 5x + 25$. Mean = $(5x + 25)/5 = x + 5 = 9 \implies x = 4$. Last three observations are $x+5, x+7, x+10$, which are $9, 11, 14$. Their mean = $(9+11+14)/3 = 34/3 = 11.33$. Wait, let's recalculate: $x+5=9, x+7=11, x+10=14$. Sum =

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