Rewrite the interval in terms of the standard deviation.
B
b
Rewrite the interval in terms of the standard deviation.
A
a
83.85 %
B
b
16 %
Practice makes perfect
a
We have been told that the times of the runners are normally distributed with a mean of μ=12.5 seconds and a standard deviation of σ=0.3 seconds. We will find the percent of runners that have a time between 11.6 seconds and 12.8 seconds.
11.6Next, we will divide the difference by the standard deviation, 0.3.
0.9/0.3=3
This means that 11.6 seconds is 3 standard deviations below the mean. We will do the same thing for the upper limit. This time, we will subtract the mean from the upper limit, 12.8 seconds.
12.8- 12.5= 0.3
Then, we will, once again, divide the difference by the standard deviation.
0.3/0.3=1
The upper limit is 1 standard deviation above the mean. As a result, we will shade the region between 3 standard deviations below and 1 standard deviation above the mean.
Finally, we can calculate the percentage of runners in the shaded region by adding the percents of the individual regions.
83.85 % of runners had a time between 11.6 seconds and 12.8 seconds.
b
This time, we will find the probability that a randomly selected runner has a time greater than 12.8 seconds and less than 13.4 seconds.
12.8above the mean. Thus, we only need to calculate how far above the mean 13.4 seconds is. To do that, we will subtract the mean from the upper limit.
13.4- 12.5= 0.9
Then, we will divide the difference by the standard deviation, 0.3.
0.9/0.3=3
The upper limit is 3 standard deviations above the mean. As a result, the region between 1 standard deviation above and 3 standard deviations above the mean will give us the probability.
Finally, we can calculate the probability by adding the percentages of the two shaded regions.
