A police radar unit is set up to monitor
vehicles on a stretch of highway with a
speed limit of 100 km/h. Long-term records
for this location show that speeds vary
normally with a mean of 105 km/h and
a standard deviation of 7 km/h. Drivers
who exceed the speed limit by 20 km/h
accumulate demerit points as well as pay
a fine.
a) What percent of the drivers will
accumulate demerit points?
b) What is the probability that a given
vehicle has a speed between 99 km/h
and 101 km/h?
Your question is perfectly suitable to use:
http://davidmlane.com/normal.html
enter 100 for mean
7 for sd
enter 120 in "above" and press enter to get .0161
I am sure you can figure out what to do for b)
Hi Reiny, Thx .... If I want to use the regular formula .....
a) Well, let's do some number crunching here. First, we need to calculate the z-score for exceeding the speed limit by 20 km/h. The formula for the z-score is (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.
So, for exceeding the speed limit by 20 km/h, the z-score would be (120 - 105) / 7 = 2.14.
Now, we can refer to a z-score table to find the corresponding percentile. In this case, a z-score of 2.14 corresponds to approximately 98.3%.
Therefore, about 98.3% of the drivers will accumulate demerit points if they exceed the speed limit by 20 km/h.
b) To find the probability that a given vehicle has a speed between 99 km/h and 101 km/h, we need to calculate the z-scores for both speeds.
For 99 km/h:
z-score = (99 - 105) / 7 = -0.86
For 101 km/h:
z-score = (101 - 105) / 7 = -0.57
Again, referring to the z-score table, we find that a z-score of -0.86 corresponds to approximately 19.74% and a z-score of -0.57 corresponds to approximately 28.08%.
To find the probability between these two speeds, we subtract the smaller probability from the larger one:
28.08% - 19.74% ≈ 8.34%
So, the probability that a given vehicle has a speed between 99 km/h and 101 km/h is approximately 8.34%. Remember, though, that it's always better to drive within the speed limit to avoid demerit points and fines!
To solve these questions, we will use the concept of the normal distribution. Given that the speeds on this stretch of highway follow a normal distribution with a mean of 105 km/h and a standard deviation of 7 km/h, we can use this information to find the requested probabilities.
a) To determine the percentage of drivers who will accumulate demerit points, we need to find the proportion of drivers whose speed exceeds the speed limit of 100 km/h.
Using the standard normal distribution table or a calculator, we can find the z-score for a speed of 100 km/h:
z = (x - μ) / σ = (100 - 105) / 7 = -0.714
Using the z-score, we can find the corresponding cumulative probability from the standard normal distribution table: P(z < -0.714) ≈ 0.238
This probability represents the proportion of drivers who will accumulate demerit points. To express this as a percentage, we multiply by 100:
0.238 * 100 = 23.8%
Therefore, approximately 23.8% of the drivers will accumulate demerit points.
b) To find the probability that a given vehicle has a speed between 99 km/h and 101 km/h, we need to find the area under the normal curve between these two speeds.
First, we need to calculate the z-scores for speeds of 99 km/h and 101 km/h:
z1 = (x1 - μ) / σ = (99 - 105) / 7 = -0.857
z2 = (x2 - μ) / σ = (101 - 105) / 7 = -0.571
Next, we use the standard normal distribution table to find the cumulative probability for each z-score:
P(z < -0.857) ≈ 0.195
P(z < -0.571) ≈ 0.283
To find the probability that a given vehicle has a speed between 99 km/h and 101 km/h, we subtract the smaller probability from the larger probability:
P(-0.857 < z < -0.571) = P(z < -0.571) - P(z < -0.857) = 0.283 - 0.195 = 0.088
Therefore, the probability that a given vehicle has a speed between 99 km/h and 101 km/h is approximately 0.088 or 8.8%.
To find the answers to these questions, we will use the concept of the standard normal distribution. Here's how we can calculate the required probabilities using the mean and standard deviation provided:
a) To find the percentage of drivers who will accumulate demerit points, we need to determine the probability of a driver exceeding the speed limit by 20 km/h or more.
First, we calculate the z-score for a speed of 120 km/h, which is 20 km/h over the speed limit. The formula for the z-score is:
z = (x - μ) / σ
where x is the value we want to find the z-score for (120 km/h), μ is the mean (105 km/h), and σ is the standard deviation (7 km/h).
Plugging these values into the formula, we get:
z = (120 - 105) / 7 = 15 / 7 = 2.14
Using a standard normal distribution table or a calculator, we can find that the cumulative probability for a z-score of 2.14 is approximately 0.9834. This means that about 98.34% of the distribution is below a speed of 120 km/h.
However, we need to find the probability of exceeding this speed, not being below it. So, we subtract this cumulative probability from 1 to get the probability of exceeding the speed limit by 20 km/h or more:
P(Exceeding speed limit by 20 km/h or more) = 1 - 0.9834 = 0.0166 = 1.66%
Therefore, approximately 1.66% of drivers will accumulate demerit points.
b) To find the probability that a given vehicle has a speed between 99 km/h and 101 km/h, we need to calculate the z-scores for both speeds and then find the probability between those z-scores.
For 99 km/h:
z1 = (99 - 105) / 7 = -6 / 7 ≈ -0.86
For 101 km/h:
z2 = (101 - 105) / 7 = -4 / 7 ≈ -0.57
Using a standard normal distribution table or a calculator, we find that the cumulative probability for a z-score of -0.86 is approximately 0.1949 and the cumulative probability for a z-score of -0.57 is approximately 0.2839.
To find the probability between these two speeds, we subtract the cumulative probability of the lower z-score from the cumulative probability of the higher z-score:
P(99 km/h < Speed < 101 km/h) = 0.2839 - 0.1949 = 0.089
Therefore, the probability that a given vehicle has a speed between 99 km/h and 101 km/h is approximately 0.089, or 8.9%.