AP Statistics 2.9 Parameters of Random Variables- Exam Style Questions - FRQs - New Syllabus

(a)
Looking at the graph, the line for vans starts higher than the line for coaches when \(p\) is small (since vans don’t lose much money even in a weak market), but the coach line rises more steeply and eventually overtakes the van line at \(p=0.76\). For any \(p\) less than \(0.76\), the van line sits above the coach line, meaning vans give the larger expected return.
We can also confirm this algebraically by plugging in a value of \(p\) less than \(0.76\), say \(p=0.5\):
Expected return for coaches:
\( 111(0.5)-27=55.5-27=28.5 \)
So the coach return is \(\$285{,}000\).
Expected return for vans:
\( 16(0.5)+45=8+45=53 \)
So the van return is \(\$530{,}000\).
Since \(53>28.5\), vans give the higher expected return at \(p=0.5\). Because both expected-return equations are linear in \(p\) and the two lines cross only once (at \(p=0.76\)), this same ordering — vans above coaches — must hold for every value of \(p\) below \(0.76\).
\( \boxed{\text{For } p<0.76, \text{ vans provide the greater expected annual return.}} \)

(b)
Since we’re estimating the proportion \(p\) of similar markets with strong demand based on a sample proportion, this calls for a one-sample \(z\)-confidence interval for a proportion.
From the sample,
\( \hat{p}=\dfrac{65}{100}=0.65 \)
Before building the interval, check the conditions. The \(100\) markets were a random sample, and since there are several thousand similar markets nationally, \(100\) is less than \(10\%\) of that population. For the large-sample condition,
\( n\hat{p}=100(0.65)=65 \)
\( n(1-\hat{p})=100(0.35)=35 \)
Both of these are well above \(10\), so a normal approximation is reasonable and the interval can proceed.
The formula for the confidence interval is:
\( \hat{p}\pm z^{*}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}} \)
Compute the standard error first:
\( \sqrt{\dfrac{(0.65)(0.35)}{100}}=\sqrt{\dfrac{0.2275}{100}}=\sqrt{0.002275} \)
\( \approx0.0477 \)
For \(95\%\) confidence, \(z^{*}=1.96\), so the margin of error is:
\( 1.96\times0.0477\approx0.0935 \)
Putting it together:
\( 0.65\pm0.0935 \)
\( \boxed{(0.5565,\ 0.7435)} \)
Interpretation: We can be \(95\%\) confident that the true proportion of similar markets that will experience strong demand is between about \(0.56\) and \(0.74\).

(c)
The interval from part (b) tells us that the plausible values for \(p\), the probability of strong demand, range from about \(0.56\) to \(0.74\). Every single value in that interval is less than \(0.76\), the cutoff found in part (a). Since part (a) showed that vans produce a larger expected return than coaches whenever \(p<0.76\), this means that for every plausible value of \(p\) suggested by the data, vans come out ahead. So choosing vans makes sense — it’s the decision that’s expected to perform better across the entire range of demand probabilities that the sample data support.

(d)
Even though the entire confidence interval lies below \(0.76\), the vice president could argue based on the actual dollar amounts at stake rather than just which option has the slightly higher expected return. Notice that throughout the confidence interval \((0.56,\ 0.74)\), the probability of strong demand is consistently estimated to be greater than \(0.5\) — meaning strong demand is more likely than weak demand in this market.

If demand turns out to be strong, the payoff table shows coaches return \(\$840{,}000\) compared to only \(\$610{,}000\) for vans — a substantial difference in the company’s favor. Meanwhile, if demand is weak, coaches lose \(\$270{,}000\) while vans still make \(\$450{,}000\), but this weak-demand scenario is the less likely one according to the interval. So the vice president could reasonably argue that, since strong demand is more probable and the upside for coaches in that case is so much larger than the modest expected-value edge vans have overall, coaches are actually the better choice for this particular route.