10 year old kids were asked to run around the track. Their times were normally distributed with a mean of 58.3 sec. and a standard deviation of 7.1 sec.
a) What percentage of times will be less than 50 seconds?
b) What percentage of times will be between 50 and 60 seconds?
To find the percentage of times that fall within a certain range, we need to use the concept of Z-scores and the standard normal distribution. The Z-score measures the number of standard deviations a given value is from the mean.
a) To find the percentage of times that will be less than 50 seconds, we need to first determine the Z-score for 50 seconds.
Z = (x - μ) / σ
Where:
x = given time (50 seconds)
μ = mean (58.3 seconds)
σ = standard deviation (7.1 seconds)
Z = (50 - 58.3) / 7.1
Z ≈ -1.17
Next, we need to look up the percentage of times falling below this Z-score using a standard normal distribution table (also known as a Z-table). The Z-table provides the area under the curve to the left of a given Z-score. In this case, we are interested in finding the area to the left of -1.17.
The Z-table shows that the area to the left of -1.17 is approximately 0.121, or 12.1%. Therefore, approximately 12.1% of the times will be less than 50 seconds.
b) To find the percentage of times that will be between 50 and 60 seconds, we can subtract the percentage from part a from the percentage of times that will be less than 60 seconds.
Using the same process as above, we find the Z-score for 60 seconds:
Z = (60 - 58.3) / 7.1
Z ≈ 0.24
Using the Z-table, we find that the area to the left of 0.24 is approximately 0.594, or 59.4%. Therefore, approximately 59.4% of the times will be less than 60 seconds.
To calculate the percentage between 50 and 60 seconds, we subtract the percentage from part a from the percentage from part b:
Percentage = 59.4% - 12.1%
Percentage ≈ 47.3%
Therefore, approximately 47.3% of the times will be between 50 and 60 seconds.