Question

In a study investigating the effect of car speed on accident severity, 5000 reports of fatal automobile accidents were examined, and the vehicle speed at impact was recorded for each one. For these 5000 accidents, the average speed was 46 mph and that the standard deviation was 15 mph. A histogram revealed that the vehicle speed at impact distribution was approximately normal. (a) Roughly what proportion of vehicle speeds were between 31 and 61 mph? (Round the answer to the nearest whole number.) Approximately % (b) Roughly what proportion of vehicle speeds exceeded 61 mph? (Round the answer to the nearest whole number.) Approximately

Answer 1

Explanation:

There is a rule called the 68-95-99.7 rule. It tells us (so we don't have to google it every time), that the area under the normal curve cut off at +/-1 standard deviation makes about 68% of the total area; +/-2 deviations correspond to 95% of such an area, and +/-3 to 99.7%.

To answer this question you only need the 68% part and use it as follows. proportion of vehicles that crash with a speed between 31 and 61 mph is exactly the area under the normal curve corresponding to a range between (mean - 1 std deviation) to (mean + 1 std deviation), or (46 - 15) to (46 + 15). And the rule tells you: that is about 68%, thus answering (a)

(b) is related to the range calculated in (a). If the area within +/- standard deviation is 68% of the total area of the normal distribution curve, then what is the complementary area? 100-68=32% This proportion corresponds to speed lower than (mean - 1 deviation) AND higher than (mean + 1 deviation). The question asks about the speeds exceeding the 61 mph mark, so that is only half of the residual area, i.e., 32/2 = 16%. So the answer for (b) is the proportion of the 62+mph crashes is 16%.

Answer 2

Following are the calculation to the given question:

Given:

$\to \bold{ mean (\mu) = 45}\\\\\to \bold{standard \ \ deviation(\sigma) = 15}$

Solve the value for A:

$P(31 <x <61) = P(\frac{(31 - 45)}{15}) < \frac{x-\mu}{\sigma} < \frac{(61 - 45)}{15}))$

                         $= P(-\frac{14}{15} < \frac{x-\mu}{\sigma} < \frac{16}{15}) \\\\= P(-0.933<z<1.066) \\\\= P(z <1.066) - P(z < -0.933)\\\\= 0.82459- 0.1432118\\\\= 0.6813782\\\\= 68.13\%$                    

Solve the value for B:

$P(x >61) = 1 - P(x <61)$

                 $= 1 - P(\frac{x-\mu}{\sigma} < \frac{(61 - 45)}{15})\\\\= 1 - P(\frac{x-\mu}{\sigma} < \frac{(16)}{15})\\\\= 1 - P(Z <1.066)\\\\= 1 - 0.1432118\\\\= 0.8567\\\\= 85.67\%$

Therefore, the final answer is "68.13% and 85.67%"

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