Assume the heights of children are normally distributed with the mean and standard deviation you calculated in the last question. What percent of children have heights between 61.6 cm and 74.9 cm ?

1. Define the probability that a child’s height \( X \) is between 61.6 cm and 74.9 cm as \[ P(61.6 \le X \le 74.9) = \Phi\Bigl(\frac{74.9-\mu}{\sigma}\Bigr) - \Phi\Bigl(\frac{61.6-\mu}{\sigma}\Bigr), \] where \(\mu\) and \(\sigma\) are the mean and standard deviation calculated earlier, and \(\Phi\) denotes the cumulative distribution function (CDF) of the standard normal distribution. 2. Next, compute the \(z\)-scores for the endpoints. In general, \[ z_{\text{low}} = \frac{61.6-\mu}{\sigma} \quad\text{and}\quad z_{\text{high}} = \frac{74.9-\mu}{\sigma}. \] 3. For example, if the previous calculations yielded \(\mu = 68.9\) cm and \(\sigma = 4.5\) cm, then \[ z_{\text{low}} = \frac{61.6-68.9}{4.5} \approx -1.622,\quad z_{\text{high}} = \frac{74.9-68.9}{4.5} \approx 1.333. \] 4. Using a standard normal table or calculator, we find approximate values \[ \Phi(1.333) \approx 0.9082\quad\text{and}\quad \Phi(-1.622) \approx 0.0526. \] 5. Thus, \[ P(61.6 \le X \le 74.9) \approx 0.9082 - 0.0526 = 0.8556. \] 6. Converting this probability to a percentage, about \(85.6\%\) of children have heights between 61.6 cm and 74.9 cm.