## **Core Concept**
The question tests understanding of statistical concepts, specifically the calculation of a confidence interval (CI) for a mean. The formula for a CI involves the mean, standard deviation (SD), and the critical value from a standard normal distribution (Z-score) for a given confidence level. For a 95% CI, the Z-score is approximately 1.96.

## **Why the Correct Answer is Right**
To calculate the 95% CI for the mean weight, we use the formula:
[ text{CI} = bar{x} pm (Z_{alpha/2} times frac{SD}{sqrt{n}}) ]
where:
- (bar{x}) is the mean weight (90 kg),
- (Z_{alpha/2}) is the Z-score for 95% confidence (1.96),
- (SD) is the standard deviation (30 kg),
- (n) is the number of students (4).

Substituting the given values:
[ text{CI} = 90 pm (1.96 times frac{30}{sqrt{4}}) ]
[ text{CI} = 90 pm (1.96 times frac{30}{2}) ]
[ text{CI} = 90 pm (1.96 times 15) ]
[ text{CI} = 90 pm 29.4 ]
So, the 95% CI is ( 90 pm 29.4 ) kg, which translates to:
[ text{CI} = (60.6, 119.4) ]

## **Why Each Wrong Option is Incorrect**
- **Option A:** This option does not match our calculated CI of ( (60.6, 119.4) ).
- **Option B:** Similarly, this does not match the calculated interval.
- **Option D:** This option also does not correspond to the calculated 95% CI.

## **Clinical Pearl / High-Yield Fact**
A key point to remember is that the width of the confidence interval is inversely proportional to the square root of the sample size (( sqrt{n} )). This means that to achieve a narrower CI (more precise estimate) with the same level of confidence, you need to increase the sample size. For 95% confidence intervals, the Z-score to use is 1.96.

## **Correct Answer:**
**Correct Answer: C. .**