The pulse rates of 165 randomly selected adult males vary from a low of 45 bpm to a high of 109 bpm . Find the minimum sample size required to estimate the mean pulse \( 90 \% \) confidence that the sample mean is within 3 bpm of the population mean. Complete parts (a) through (c) below. a. Find the sample size using the range rule of thumb to estimate \( \sigma \). \( \mathrm{n}=77 \) (Round up to the nearest whole number as needed.) b. Assume that \( \sigma=11.7 \mathrm{bpm} \), based on the value \( \mathrm{s}=11.7 \mathrm{bpm} \) from the sample of 165 male pulse rates. \( \mathrm{n}=\square \) (Round up to the nearest whole number as needed.)

To find the minimum sample size required for estimating the mean pulse rate with a specified confidence level and margin of error, we can use the formula: \[ n = \left(\frac{Z \cdot \sigma}{E}\right)^2 \] where \( Z \) is the Z-score corresponding to the desired confidence level, \( \sigma \) is the standard deviation, and \( E \) is the margin of error. For a 90% confidence level, the Z-score is approximately 1.645. **a.** Using the range rule of thumb to estimate \( \sigma \): The range is \( 109 - 45 = 64 \). The range rule of thumb suggests that \( \sigma \approx \frac{Range}{4} = \frac{64}{4} = 16 \). Now substituting in the formula: \[ n = \left(\frac{1.645 \cdot 16}{3}\right)^2 \approx \left(\frac{26.32}{3}\right)^2 \approx (8.77)^2 \approx 77.04 \] So, rounding up gives you \( n = 78 \). **b.** Assuming \( \sigma = 11.7 \, bpm \): Using the same formula: \[ n = \left(\frac{1.645 \cdot 11.7}{3}\right)^2 \approx \left(\frac{19.24}{3}\right)^2 \approx (6.41)^2 \approx 41.00 \] Rounding up gives you \( n = 41 \). So, your final answers are: a. \( n = 78 \) b. \( n = 41 \)