Question

The average marks of 15 students in a subject is 65. If another student has scored 97 marks in the same subject, then the average marks of the 16 students is:

• A. 69
• B. 71
• C. 68
• D. 67

Hint:

Using the deviation technique saves time and keeps numbers manageable: $$\text{New Average} = \text{Old Average} + \frac{\text{New Score} - \text{Old Average}}{\text{New Total Count}}$$ $$\text{New Average} = 65 + \frac{97 - 65}{16} = 65 + \frac{32}{16} = 65 + 2 = 68$$

Answer 1

The Correct Option is C

Concept: The statistical mean or average of a dataset is defined as the total sum of all individual observations divided by the absolute number of observations present within that set. $$\text{Average } (\mu) = \frac{\text{Sum of all observations}}{\text{Total number of observations}} \quad \Rightarrow \quad \text{Sum} = \text{Average} \times \text{Count}$$

Step-by-step Explanation: Method A: Traditional Cumulative Calculation

Step 1: Compute the initial aggregate score for the first group.
We are given that the sample size $n_1 = 15$ students has a mean score of $\mu_1 = 65$. $$\text{Sum}_{15} = \text{Average}_{15} \times 15$$ $$\text{Sum}_{15} = 65 \times 15 = 975$$

Step 2: Incorporate the incoming data point to find the new total score.
A 16th student joins the dataset with a verified individual score of 97 marks. We add this value to our previous sum: $$\text{Sum}_{16} = \text{Sum}_{15} + \text{Score}_{\text{new}}$$ $$\text{Sum}_{16} = 975 + 97 = 1072$$

Step 3: Divide the updated sum by the total count of observations.
The total count of students has increased to $n_2 = 16$. The new mean score ($\mu_2$) is: $$\text{Average}_{16} = \frac{\text{Sum}_{16}}{16} = \frac{1072}{16}$$ Executing long division: $$1072 \div 16 = 68$$ Thus, the new mean mark across all 16 students is 68. Method B: Deviational Analysis (Shortcut Method)
Alternatively, we can analyze the surplus points the new student introduces relative to the existing average baseline: $$\text{Baseline Average} = 65$$ $$\text{New Student's Score} = 97$$ $$\text{Surplus Score } (\Delta) = 97 - 65 = +32 \text{ marks}$$ This surplus of 32 marks must be distributed equally among all 16 students in the expanded group: $$\text{Average Increase per student} = \frac{32 \text{ marks}}{16 \text{ students}} = +2 \text{ marks}$$ $$\text{New Average} = \text{Old Average} + \text{Increase} = 65 + 2 = 68$$