Given readings

The measured time periods are: $4.62\,\text{s},\; 4.632\,\text{s},\; 4.6\,\text{s},\; 4.64\,\text{s}$

Find the arithmetic mean

$\bar{T}=\frac{4.62+4.632+4.6+4.64}{4}$

First add: $4.62+4.632+4.6+4.64 = 18.492$

Now divide by $4$: $\bar{T}=\frac{18.492}{4}=4.623\,\text{s}$

Apply the rule of significant figures

The readings have different precisions:

• $4.632\,\text{s}$ is measured up to $0.001\,\text{s}$
• $4.62\,\text{s}$ and $4.64\,\text{s}$ are measured up to $0.01\,\text{s}$
• $4.6\,\text{s}$ is measured only up to $0.1\,\text{s}$

Since the least precise measurement is $4.6\,\text{s}$, the final mean should be reported only up to one decimal place.

So, $4.623\,\text{s} \approx 4.6\,\text{s}$

Check options

• A: $5\,\text{s}$ ✗
• B: $4.6\,\text{s}$ ✓
• C: $4.62\,\text{s}$ ✗
• D: $4.623\,\text{s}$ ✗

Final answer

The arithmetic mean in correct significant figure is: $\boxed{4.6\,\text{s}}$