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 ?

To answer this question, we first need to calculate the z-scores for both heights using the formula \( z = \frac{(X - \mu)}{\sigma} \), where \( \mu \) is the mean, \( \sigma \) is the standard deviation, and \( X \) is the value for which we want to find the z-score. Once we have the z-scores, we can use the standard normal distribution table (or a calculator) to find the corresponding percentiles and calculate the percentage of children within that height range. To illustrate, let’s say the mean height (μ) is 68 cm, and the standard deviation (σ) is 6 cm. For 61.6 cm: \( z_1 = \frac{(61.6 - 68)}{6} \approx -1.07 \) For 74.9 cm: \( z_2 = \frac{(74.9 - 68)}{6} \approx 1.83 \) Now, looking up these z-scores in the standard normal distribution table or using a cumulative distribution function calculator, we get approximately 14.5% for \( z_1 \) and 96.4% for \( z_2 \). By subtracting these two percentages, we find that about 81.9% of children have heights between 61.6 cm and 74.9 cm. So, there you have it – a fun journey through the fascinating world of statistics!