Question 699044
1) The mean number of drivers in a sample of 1000 drivers that never use a cell phone when driving is equal to n*p = 1000*.15 = 150. The standard deviation of a single trial, that is, a trial that returns 1 if the driver doesn't use a cell phone and 0 if he does is {{{sqrt(p*(1-p)) = sqrt((.15)(.85)) = .35707. You can find a Z value from the Central Limit Theorem with the formula:<br>

{{{Z = (X - mu)/(sigma*sqrt(n))}}}, where X is the number of drivers that is the "cutoff" (here, it would be 130), mu is the mean, which is 150, sigma is the standard deviation, which is .35707, and n is the total number of trials, which is 1000.<br>

However, since the binomial distribution is a discrete distribution and the normal distribution is continuous, you may need to use the continuity correction in this problem. Here, since we want to know the probability that the number of drivers who never use a cell phone is > 130, we'd adjust this by .5 and use X = 129.5 instead of 130 in the calculation. Assuming the continuity correction is required, the Z-value is:<br>

{{{Z = (129.5 - 150)/(.35707*sqrt(1000)) = -1.8155}}}<br>

Looking this value up on a Z-table such as the one at http://lilt.ilstu.edu/dasacke/eco148/ztable.htm, we find Z(-1.82) [most Z-tables round to the nearest hundredth] = .0344. The value on the Z-table is the probability that the Z value found is less than the given value, though -- we want the probability that Z > -1.82, or that n >= 130. This probability is 1 - .0344 = .9656.<br>

Note that some instructors do not use the continuity correction. If not, re-do the calculation above with X = 130 instead of 129.5.<br>

2) The probability that the third driver will be the first one who never uses a cell phone is the probability that the first two drivers do use cell phones times the probability that the third driver does not. The probability that a given driver uses a cell phone is 1 - .15 = .85, since the event that using a cell phone while driving is the compliment of never using a cell phone when driving. Thus, the probability that the third driver is the first driver to never use a cell phone when driving is<br>

(.85)*(.85)*(.15) = .108375, or .1084 to 4 decimal places.<br>

3) The random variable that is the number of drivers required to find the first who does not use a cell phone when driving is geometrically distributed with p  = .15. The expected value of the geometric distribution is equal to 1/p, which is 1/.15 = 20/3. The nearest whole number is 7, so they can expect to check 7 drivers before finding the first who never uses a cell phone.<br>

4) I'm not sure if you meant the first driver who does not use a cell phone, because in the question you ask for the first who does. I'm going to work the problem as if they want the first who does not. The expected cost is the cost per driver times the expected number of drivers required to find the first driver who does not use a cell phone. This is 5*(20/3) = $33.33. The variance of this distribution, assuming that it only costs 1 dollar to question someone instead of five, is<br>

{{{(1-p)/(p^2) = 37.77778}}}<br>

However, it costs $5 to question a driver. The random variable is now five times the value given by a geometric distribution. The rule for scaling the variance of a distribution by a constant is<br>

{{{Var(aX) = a^2*Var(X)}}}<br>

So the variance of questioning the drivers if it costs $5 per driver is 5*5*37.77778 = 944.44444. The standard deviation is the square root of this value, which is 30.7318.<br>

If the question actually does ask for the first driver who does use a cell phone instead of the first that does not, then repeat the calculations with p = .85 instead of p = .15.