A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 44 specimens and counts the number of seeds in each. Use her sample results (mean $=47.7$, standard deviation $=7.7$ ) to find the $98 \%$ confidence interval for the number of seeds for the species. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. \[ \text { 98\% C.I. }= \]

Step 1 ：We are given that the botanist has sampled 44 specimens of a certain fruit and counted the number of seeds in each. The mean number of seeds is 47.7 and the standard deviation is 7.7. We are asked to find the 98% confidence interval for the number of seeds for the species.

Step 2 ：The formula for the confidence interval is \(\text{mean} \pm \left( Z \times \frac{\text{standard deviation}}{\sqrt{\text{sample size}}} \right)\), where Z is the Z-score corresponding to the desired confidence level.

Step 3 ：For a 98% confidence level, the Z-score is approximately 2.33. This value can be found from a Z-table or using a standard normal distribution.

Step 4 ：Substituting the given values into the formula, we get \(47.7 \pm \left( 2.33 \times \frac{7.7}{\sqrt{44}} \right)\).

Step 5 ：Solving the above expression, we get the margin of error as approximately 2.705.

Step 6 ：Subtracting this margin of error from the mean, we get the lower bound of the confidence interval as approximately 44.995.

Step 7 ：Adding the margin of error to the mean, we get the upper bound of the confidence interval as approximately 50.405.

Step 8 ：\(\boxed{\text{Therefore, the 98% confidence interval for the number of seeds for the species is approximately } (44.995, 50.405)}\)