Two students `A` & `B` carried out an experiment to measure the time period of a pendulum which in actual is `5.40s`. Their experimental readings are as follows`:`
`{:(StudentA," ",StudentB,),(5.43s," ",5.32s,),(5.37s," ",5.34s,),(5.41s," ",5.30s,),(5.52s," ",5.29s,),(5.51s," ",5.33s,):}` Then

Options

A. Student `A` is more precise but less accurate
B. Student `B` is more precise but less accurate
C. Students `A` is more precise as well as accurate
D. Students `B` is more precise as well as accurate

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the experimental readings
We have the experimental readings for both students A and B:
- **Student A's readings**: 5.43 s, 5.37 s, 5.41 s, 5.52 s, 5.51 s
- **Student B's readings**: 5.32 s, 5.34 s, 5.30 s, 5.29 s, 5.33 s

### Step 2: Calculate the average (mean) value for each student
To find the average value, we will sum the readings and divide by the number of readings (5).

**For Student A:**
\[ \text{Average}_A = \frac{5.43 + 5.37 + 5.41 + 5.52 + 5.51}{5} \]
Calculating this gives:
\[ \text{Average}_A = \frac{27.24}{5} = 5.448 \text{ s} \]

**For Student B:**
\[ \text{Average}_B = \frac{5.32 + 5.34 + 5.30 + 5.29 + 5.33}{5} \]
Calculating this gives:
\[ \text{Average}_B = \frac{26.58}{5} = 5.316 \text{ s} \]

### Step 3: Determine the maximum deviation for each student
The maximum deviation is found by subtracting the minimum reading from the maximum reading.

**For Student A:**
- Maximum reading = 5.52 s
- Minimum reading = 5.37 s
\[ \text{Max Deviation}_A = 5.52 - 5.37 = 0.15 \text{ s} \]

**For Student B:**
- Maximum reading = 5.34 s
- Minimum reading = 5.29 s
\[ \text{Max Deviation}_B = 5.34 - 5.29 = 0.05 \text{ s} \]

### Step 4: Compare precision and accuracy
- **Precision** refers to how close the measurements are to each other. Student B has a smaller maximum deviation (0.05 s) compared to Student A (0.15 s), indicating that Student B is more precise.
- **Accuracy** refers to how close the average measurement is to the actual value (5.40 s).

**Comparing averages to the actual value:**
- Average of A = 5.448 s (deviation from actual = |5.448 - 5.40| = 0.048 s)
- Average of B = 5.316 s (deviation from actual = |5.316 - 5.40| = 0.084 s)

Since Student A's average is closer to the actual value, Student A is more accurate.

### Conclusion
- **Student A** is more accurate.
- **Student B** is more precise.

### Final Answer
Student B is more precise but less accurate than Student A.

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