at one high school the mean time for running the 100-yard dash is 15.2 seconds with a standard deviation of 0.9 seconds. The times are very close approximated by a normal curve. Find the percent of times that are:
between 15.2 and 16.1 seconds
To find the percent of times that are between 15.2 and 16.1 seconds, we can use the standard normal distribution and the z-score formula.
1. Calculate the z-scores for the given times:
The z-score formula is given by:
z = (x - μ) / σ
where x is the given time, μ is the mean time, and σ is the standard deviation.
For the lower time, x = 15.2 seconds:
z1 = (15.2 - 15.2) / 0.9 = 0
For the upper time, x = 16.1 seconds:
z2 = (16.1 - 15.2) / 0.9 ≈ 1
2. Find the area between the z-scores using a standard normal distribution table or a calculator.
Using either method, we find that the area between z = 0 and z ≈ 1 is approximately 0.3413.
3. Convert the area to a percentage:
To find the percentage, multiply the decimal by 100:
0.3413 * 100 ≈ 34.13%
Therefore, approximately 34.13% of times are between 15.2 and 16.1 seconds.
To find the percent of times that are between 15.2 and 16.1 seconds, we need to calculate the z-scores for these two values and find the area under the normal curve between these two z-scores.
Step 1: Calculate the z-score for 15.2 seconds.
The z-score formula is:
z = (x - μ) / σ
Where:
x = the given value (15.2 seconds)
μ = mean (15.2 seconds)
σ = standard deviation (0.9 seconds)
Using the formula:
z = (15.2 - 15.2) / 0.9
z = 0
Step 2: Calculate the z-score for 16.1 seconds.
Using the same formula:
z = (16.1 - 15.2) / 0.9
z = 1
Step 3: Find the area under the normal curve between these two z-scores.
To find the area, we can use a standard normal distribution table or a statistical calculator.
Looking up the z-scores in a standard normal distribution table, we find:
The area to the left of z = 0 is 0.5000
The area to the left of z = 1 is 0.8413
To find the area between z = 0 and z = 1, we subtract the area to the left of z = 0 from the area to the left of z = 1:
0.8413 - 0.5000 = 0.3413
Therefore, approximately 34.13% of times are between 15.2 and 16.1 seconds at this high school.
I suggest you learn how to use (Broken Link Removed) or tables of the Gaussian error funtion.
The answer that I get using the computational tool at that site is 34.1%