Suppose that diastolic blood pressure readings of adult males have a bell-shaped distribution with a mean of 8080 mmHg and a standard deviation of 1111 mmHg. Using the empirical rule, what percentage of adult males have diastolic blood pressure readings that are at least 6969 mmHg? Please do not round your answer.

84% of adult males have diastolic blood pressure readings that are at least 69 mmHg

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 = 80

Standard deviation = 11

Using the empirical rule, what percentage of adult males have diastolic blood pressure readings that are at least 69 mmHg?

The normal probability distribution is symmetric, which means that 50% of the measures are above the mean and 50% are below.

69 = 80 - 11

So 69 is one standard deviation below the mean.

By the Empirical Rule, 68% of the measures below the mean, are within 1 standard of the mean. So

50% + 0.68*50% = 84%

84% of adult males have diastolic blood pressure readings that are at least 69 mmHg

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.