Question

1. driving test you are about to take the road test for your drivers license. you hear that only 34% of candidates pass the test the first time, but the percentage rises to 72% on subsequent retests. because most teenagers really want to drive, they keep taking the test until they pass!

a) create a plan for a simulation to estimate the average number of tests drivers take in order to get a license.
b) the histogram in the next column shows the results of 100 trials of a simulation. use these results to estimate the average number of tests drivers take in order to get a license.

Explanation:

Step1: Simulation Plan (for part a)

• Assign digits: Let \(00 - 33\) represent passing on the first test (since \(34\%\) of candidates pass first - time, and we use two - digit random numbers \(00 - 99\)). For subsequent tests, let \(00 - 71\) represent passing (since \(72\%\) pass on retests).
• Generate numbers: Use a random number generator (e.g., in a spreadsheet or a random number table).
• Conduct trials: For each trial, start by generating a random two - digit number. If it is in the \(00 - 33\) range, the number of tests is \(1\). If not, for the second test, generate another random two - digit number. If it is in the \(00 - 71\) range, the number of tests is \(2\). If not, continue generating numbers for each subsequent test until a number in the appropriate range (for that test) is generated.
• Repeat: Repeat the process for a large number of trials (say \(N\) trials).

Step2: Calculate average (for part b)

Let \(x_i\) be the number of tests in the \(i -\)th trial (\(i = 1,2,\cdots,100\)). The formula for the average \(\bar{x}=\frac{\sum_{i = 1}^{100}x_i}{100}\).
Assume from the histogram (not shown here, but using the general formula for the mean of a frequency distribution): If we have frequencies \(f_1,f_2,\cdots,f_k\) for values \(x_1,x_2,\cdots,x_k\) (where \(x_j\) is the number of tests and \(f_j\) is the frequency of that number of tests in the \(100\) trials), then \(\bar{x}=\frac{\sum_{j = 1}^{k}x_jf_j}{100}\). For example, if \(x_1 = 1\) with \(f_1=34\) (corresponding to the \(34\%\) first - time pass in the simulation), \(x_2 = 2\) with \(f_2=(100 - 34)\times0.72\approx48.72\approx49\) (since \(66\%\) take a second test and \(72\%\) of them pass), \(x_3=3\) with \(f_3=(100 - 34 - 49)=17\) (by subtraction). Then \(\bar{x}=\frac{1\times34 + 2\times49+3\times17}{100}=\frac{34 + 98+51}{100}=\frac{183}{100}=1.83\)

Answer:

a) The simulation plan is as described above.
b) The average number of tests (using the example calculation) is \(1.83\). The actual value depends on the frequencies from the histogram. If we assume the frequencies from the first - time pass proportion and retest proportion: \(\text{Average number of tests}\approx1.8\) (rounded to one decimal place).