To calculate the 95% confidence interval for the mean score of all golfers, we can use the formula for the confidence interval when the population standard deviation is known:
\[ \text{Confidence Interval} = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \]
Where:
- \( \bar{x} \) is the sample mean.
- \( z \) is the z-score corresponding to the desired confidence level (for 95%, \( z \approx 1.96 \)).
- \( \sigma \) is the population standard deviation.
- \( n \) is the sample size.
Given:
- \( \bar{x} = 92 \)
- \( \sigma = 5 \)
- \( n = 35 \)
Step 1: Calculate the standard error (SE):
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{35}} \approx \frac{5}{5.916} \approx 0.845 \]
Step 2: Calculate the margin of error (ME):
\[ ME = z \times SE = 1.96 \times 0.845 \approx 1.655 \]
Step 3: Calculate the confidence interval:
\[ \text{Lower Limit} = \bar{x} - ME \approx 92 - 1.655 \approx 90.345 \] \[ \text{Upper Limit} = \bar{x} + ME \approx 92 + 1.655 \approx 93.655 \]
Step 4: Round to the nearest whole numbers:
- Lower Limit: \( \approx 90.345 \) rounds to 90
- Upper Limit: \( \approx 93.655 \) rounds to 94
Thus, the 95% confidence interval for the mean score is:
\[ (90, 94) \]
So the final answer is:
(90,94)