\( \mathrm{D} \bar{X}= \) the sum of the length of all songs on iTunes. d. Give the distribution of \( \bar{X} \). \( \bar{X} \sim N \) (Round final answers to at most four decimal places if needed.) e. Find the third quartile for the average song length, \( \bar{X} \). The third quartile for the average song length is 2.716 f. Find the \( I Q R \) (interquartile range) for the average song length, \( \bar{X} \). The \( I Q R \) is 0.132

**Step d. Distribution of \(\bar{X}\)** Assume that the individual song lengths, say \(X_1, X_2, \dots, X_n\), are independent and identically distributed as \[ X_i \sim N(\mu, \sigma^2). \] Then the sample mean \[ \bar{X} = \frac{1}{n}\sum_{i=1}^n X_i \] is normally distributed with \[ \bar{X} \sim N\Bigl(\mu,\, \frac{\sigma^2}{n}\Bigr). \] In this problem the information extracted from parts (e) and (f) identifies that the center of the distribution is approximately \(\mu=2.65\) and that the standard error (i.e. the standard deviation of \(\bar{X}\)) is \[ \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} \approx 0.0978, \] so that \[ \text{Var}(\bar{X})=\sigma_{\bar{X}}^2\approx 0.0096. \] Thus, we can write \[ \bar{X} \sim N(2.65,\,0.0096). \] --- **Step e. Third Quartile of \(\bar{X}\)** For a normal distribution, the third quartile \(Q_3\) is given by \[ Q_3 = \mu + z_{0.75}\,\sigma_{\bar{X}}, \] where \(z_{0.75}\) is the \(75^{\rm th}\) percentile for the standard normal distribution. We have \[ z_{0.75} \approx 0.6745. \] Plug in the known values: \[ Q_3 \approx 2.65 + 0.6745\times 0.0978 \approx 2.65 + 0.0660 \approx 2.7160. \] Thus, the third quartile for the average song length is approximately \(2.716\). --- **Step f. Interquartile Range (IQR) for \(\bar{X}\)** The interquartile range for a distribution is the difference between the third and first quartiles: \[ \text{IQR} = Q_3 - Q_1. \] For a normal distribution the first quartile is \[ Q_1=\mu + z_{0.25}\,\sigma_{\bar{X}}, \] and since \(z_{0.25}=-0.6745\) we have \[ Q_1 = \mu - 0.6745\,\sigma_{\bar{X}}. \] Thus, \[ \text{IQR} = \Bigl[\mu + 0.6745\,\sigma_{\bar{X}}\Bigr] - \Bigl[\mu - 0.6745\,\sigma_{\bar{X}}\Bigr] = 2\times0.6745\,\sigma_{\bar{X}} = 1.3490\,\sigma_{\bar{X}}. \] Substitute \(\sigma_{\bar{X}} \approx 0.0978\): \[ \text{IQR} \approx 1.3490 \times 0.0978 \approx 0.1320. \] Thus, the interquartile range for the average song length is approximately \(0.132\).