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In many states a motorist is legally drunk or driving under influence (DUI), if

ID: 3126573 • Letter: I

Question

In many states a motorist is legally drunk or driving under influence (DUI), if his or her blood alcohol concentration, Y, is 0.10% or higher. When a suspected DUI offender is pulled over, police often request a sobriety test. Although the breath analyzers used for that purpose are remarkably precise, the machines exhibit a certain amount of measurement error. Because of that variability, the possibility exists that a driver's true blood alcohol concentration may be under 0.10% even though the analyzer gives a reading over 0.10%. Experience shows that repeated breath analyzer measurements taken on the same individual produce a distribution of responses that can be described by a normal pdf with mu equal to the person's true blood alcohol concentration and sigma = 0.004%. Suppose a driver is stopped at a roadblock on his way home from a party. Having celebrated a bit more than he should have, he has a true blood alcohol concentration of 0.095%, just barely below the legal limit. If he takes a breath analyzer test, what are the chances that he will be incorrectly booked on a DUI charge? Suppose the driver's blood alcohol concentration is actually 0.12% rather than 0.095%. What is the probability that the breath analyzer will make an error in his favor and indicate that he is not legally drunk? Suppose the police offer the driver a choice - either take the sobriety test once or take it twice and average the reading? Which option should a "0. 90 %" driver take? Which option should a "0.11%" driver take?

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.1      
u = mean =    0.095      
          
s = standard deviation =    0.004      
          
Thus,          
          
z = (x - u) / s =    1.25      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.25   ) =    0.105649774 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.1      
u = mean =    0.12      
          
s = standard deviation =    0.004      
          
Thus,          
          
z = (x - u) / s =    -5      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -5   ) =    2.86652*10^-7 [ANSWER]

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c)

The 0.09% driver should take it twice, as that makes it less probable that he is guilty due to less variation.

The 0.11% driver should take it once, as that makes it less probable that he is guilty due to more variation.

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