The speed limit outside a school is 40 km / h. Year 11 students measured the speed of passing vehicles over a period of time. They found the set of data to be normally distributed with a mean speed of 36 km / h and a standard deviation of 2 km / h.
What percentage of the vehicles passed the school at a speed greater than 40 km / h?
The answer is 2.5%, though I am not sure how to get this.
z-score for your event = (40 - 36)/2 = 2
I don't know if you use tables or some other means to find
the answer.
If you have tables, find 2.0 in the left column and you should see
.9772 or 97.72%
This is the percentage that would have below 40 km/h
So the percentage above 40 km/h would be 100% - 97.72^ = 2.28%
I got my results here:
http://davidmlane.com/normal.html
To find the percentage of vehicles that passed the school at a speed greater than 40 km/h, we can use the concept of the standard normal distribution.
The first step is to standardize the value of 40 km/h using the given information of the mean and standard deviation. We can do this by calculating the z-score.
Z-Score Formula:
z = (x - μ) / σ
where:
z = z-score
x = observed value
μ = mean
σ = standard deviation
In this case, x = 40 km/h, μ = 36 km/h, and σ = 2 km/h.
z = (40 - 36) / 2
z = 4 / 2
z = 2
Now that we have the z-score, we can use a standard normal distribution table (also known as a z-table) to find the area under the curve to the right of the z-score.
Looking up the z-score of 2 in the z-table, we find that the area to the right of z = 2 is approximately 0.0228. This means that about 2.28% of the vehicles passed the school at a speed greater than 40 km/h.
However, the question asks for the percentage of vehicles that passed at a speed greater than 40 km/h, which includes the area to the right of 40 km/h. Since the standard normal distribution is symmetric, we can subtract the value we found from 0.5 to get the desired percentage.
Percentage = 0.5 - 0.0228 = 0.4772
Finally, we convert this decimal value to a percentage:
Percentage = 0.4772 * 100% = 47.72%
Therefore, approximately 47.72% of the vehicles passed the school at a speed greater than 40 km/h.
To find the percentage of vehicles that passed the school at a speed greater than 40 km/h, we can use the standard normal distribution table (also known as the z-table) to calculate the z-score and then find the corresponding percentage.
First, we need to convert the given speed of 40 km/h into a z-score. The z-score formula is given by:
z = (x - μ) / σ
where:
x = the observed value (40 km/h)
μ = the mean (36 km/h)
σ = the standard deviation (2 km/h)
Plugging in the values:
z = (40 - 36) / 2
z = 4 / 2
z = 2
Now, we need to find the percentage of observations that lie to the right of this z-score on the standard normal distribution curve. Looking at the z-table, we find that the percentage corresponding to a z-score of 2 is approximately 0.9772.
However, we want to find the percentage of vehicles that passed the school at a speed greater than 40 km/h, which is the area to the right of the z-score. Since the total percentage under the curve is 1, we subtract 0.9772 from 1:
1 - 0.9772 = 0.0228 or 2.28%
Therefore, approximately 2.28% (rounded to 2.5%) of the vehicles passed the school at a speed greater than 40 km/h.