A random sample of adult female reaction times has a sample mean of x¯=394.3 milliseconds and sample standard deviation of s=84.6 milliseconds. Use the Empirical Rule to determine the approximate percentage of adult female reaction times that lie between 140.5 and 648.1 milliseconds.

The percentage of adult female reaction times that lie between 140.5 and 648.1 milliseconds is 99.7%.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 394.3 ms

Standard deviation = 84.6 ms

Use the Empirical Rule to determine the approximate percentage of adult female reaction times that lie between 140.5 and 648.1 milliseconds.

140.5 = 394.3 - 3*84.6

So 140.5 is 3 standard deviations below the mean.

648.1 = 394.3 + 3*84.6

So 648.1 is 3 standard deviations above the mean.

By the Empirical Rule,

The percentage of adult female reaction times that lie between 140.5 and 648.1 milliseconds is 99.7%.

step-by-step explanation:

the true statements are:

r = set of real numbers

n = set of natural numbers, 1,2,

w = n u{0} = {0,1,}