Question

use the image to answer the following question
the empirical rule
given a dataset of 100 values with the normal model ( n(71, 5) ), approximately how many values should fall between 61 and 71 for the normal model to hold? 11 points
48
34
20
16

Explanation:

Step1: Analyze the normal distribution parameters

Given the normal model \(N(71,6)\), where \(\mu = 71\) (mean) and \(\sigma=6\). We want to find the number of values between \(61\) and \(71\). Calculate \(71 - 61=10\), and \(\frac{10}{6}\approx1.67\). But using the empirical - rule approximations (since \(61 = 71- 10\approx71 - 2\times5\) is not exact, but we can also note that \(61=71 - 10\) and if we consider the empirical rule's \(1\sigma\) and \(2\sigma\) intervals. However, more accurately, \(61 = 71-10\) and since \(10 = \frac{10}{6}\sigma\approx1.67\sigma\). But if we use the fact that \(61 = 71 - 2\times5\) (a wrong - way of looking, better: \(61=\mu - \frac{10}{6}\sigma\). But using the empirical rule's \(1\sigma\) and \(2\sigma\) intervals. Another way: \(61 = 71-10\) and for a normal distribution \(N(\mu,\sigma)\), \(x_1=\mu - k\sigma\). Here \(k=\frac{71 - 61}{6}=\frac{10}{6}\approx1.67\). But using the empirical rule's \(1\sigma\) (\(68\%\) within \(\mu\pm\sigma\)) and \(2\sigma\) (\(95\%\) within \(\mu\pm2\sigma\)) approximations. The value \(61\) is \(\mu - \frac{10}{6}\sigma\approx\mu - 1.67\sigma\). But if we consider the left - hand side of the mean. The proportion of data within \(\mu - \sigma\) and \(\mu\) is \(34\%\), and within \(\mu - 2\sigma\) and \(\mu-\sigma\) is \(13.5\%\). The value \(61\) is \(\mu-(71 - 61)\) and \(71-61 = 10\), \(\sigma = 6\). \(61=\mu-\frac{10}{6}\sigma\approx\mu - 1.67\sigma\). Using the empirical rule's \(1\sigma\) and \(2\sigma\) intervals (a rough approximation). The proportion of data between \(61\) and \(71\) is approximately \(34\%+13.5\% = 47.5\%\) (if we consider \(61\) as \(\mu - 2\sigma\) (since \(71-2\times6=71 - 12 = 59\) is close to \(61\) in the context of the empirical - rule - based multiple - choice).

Step2: Calculate the number of values

We have \(n = 60\) values. The number of values \(N\) between \(61\) and \(71\) is \(N=n\times p\), where \(p\) is the proportion of data in the interval. If we use the empirical rule's \(1\sigma\) and \(2\sigma\) approximations (a bit of a stretch, but for multiple - choice). The proportion of data in the interval (using \(61\approx\mu - 2\sigma\) (since \(71-2\times6 = 59\) and \(61\) is close in the multiple - choice context) from the left - hand side of the mean). The proportion of data from \(\mu - 2\sigma\) to \(\mu\) is \(13.5\%+34\%=47.5\%\). So \(N = 60\times0.475 = 28.5\approx29\)

Answer:

\(29\)