For a population with \( \mu=25 \) and \( \sigma=5 \), we would expect \( 90 \% \) of all \( \bar{x} \) 's calculated from \( n=35 \) to fall between \( \square \) and \( \square \) (Round to two decimals.)

To find the range of values within which 90% of all sample means (\( \bar{x} \)) calculated from a sample size of \( n=35 \) are expected to fall, we can use the properties of the normal distribution. Given: - Population mean (\( \mu \)) = 25 - Population standard deviation (\( \sigma \)) = 5 - Sample size (\( n \)) = 35 - Confidence level = 90% The formula to find the range of values within which a certain percentage of sample means fall is given by: \[ \mu - z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} \leq \bar{x} \leq \mu + z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} \] where: - \( z_{\alpha/2} \) is the z-score corresponding to the desired confidence level - \( \mu \) is the population mean - \( \sigma \) is the population standard deviation - \( n \) is the sample size First, we need to find the z-score corresponding to a 90% confidence level. The z-score for a 90% confidence level is approximately 1.645. Now, we can substitute the given values into the formula to find the range of values within which 90% of all sample means are expected to fall. The lower limit of the range is approximately 23.61. Next, we calculate the upper limit of the range using the same formula. The upper limit of the range is approximately 26.39. Therefore, we would expect 90% of all sample means calculated from a sample size of 35 to fall between approximately 23.61 and 26.39.