d. For the 18 people, find the probability that the average coffee consumption is between 15.5 and 17.5 ounces of coffee per day. e. For part d), is the assumption that the distribution is normal necessary? ONOO Yes f. Find the IQR for the average of 18 coffee drinkers. Q1 =| Q3 = ounces ounces IQR: ounces

d.

Let $\overline{x}$ denotes the average coffee consumption.

The average coffee consumption for 18 people is,

${\mu }_{\overline{x}}=16$

The standard deviation coffee consumption for 18 people is,

$\begin{array}{rcl}{\sigma }_{\overline{x}}& =& \frac{\sigma }{\sqrt{n}}\\ & =& \frac{4}{\sqrt{18}}\\ & =& 0.9428\end{array}$

The probability that the average coffee consumption is between 15.5 and 17.5 ounces of coffee per day is,

$\begin{array}{rcl}P\left(15.5\le \overline{x}\le 17.5\right)& =& P\left(\frac{15.5-16}{0.9428}\le \frac{\overline{x}-16}{0.9428}\le \frac{17.5-16}{0.9428}\right)\\ & =& P\left(\frac{-0.5}{0.9428}\le z\le \frac{1.5}{0.9428}\right)\\ & =& P\left(-0.5303\le z\le 1.5910\right)\\ & =& P\left(z\le 1.5910\right)-P\left(z\le -0.5303\right)\end{array}$

The probability of z less than -0.5303 can be obtained using the excel formula “=NORM.S.DIST(-0.5303,TRUE)”. The probability value is 0.2980.

The probability of z less than 1.5910 can be obtained using the excel formula “=NORM.S.DIST(1.5910,TRUE)”. The probability value is 0.9442.

The required probability is,

$\begin{array}{rcl}P\left(15.5\le \overline{x}\le 17.5\right)& =& P\left(z\le 1.5910\right)-P\left(z\le -0.5303\right)\\ & =& 0.9442-0.2980\\ & =& 0.6462\end{array}$

Thus, the probability that the average coffee consumption is between 15.5 and 17.5 ounces of coffee per day is 0.6462.