Final answer:
The proportion of individuals in the population with a resting heart rate exceeding 85 is approximately 0.1587. The probability of at least two individuals in a sample of size 10 having a resting heart rate exceeding 85 is approximately 0.4871.
Step-by-step explanation:
To answer question a, we can use the standard normal distribution table to find the z-score corresponding to a resting heart rate of 85. The formula to calculate the z-score is (x - mean) / standard deviation, where x is the resting heart rate, mean is the mean resting heart rate in the population, and standard deviation is the standard deviation of resting heart rates in the population. Plugging in the values:
(85 - 80) / 5 = 1
Looking up the z-score of 1 in the standard normal distribution table, we find that the area to the left of the z-score is approximately 0.8413. To find the proportion of individuals with a resting heart rate exceeding 85, we subtract this area from 1:
1 - 0.8413 = 0.1587
Therefore, the proportion of individuals in the population with a resting heart rate exceeding 85 is approximately 0.1587.
To answer question b, we can use the binomial probability formula. The probability of at least two individuals in a sample of size 10 having a resting heart rate exceeding 85 can be calculated as:
P(X ≥ 2) = 1 - P(X = 0) - P(X = 1)
where X follows a binomial distribution with parameters n = 10 (sample size) and p = 0.1587 (probability of an individual having a resting heart rate exceeding 85 in the population). Using a binomial probability calculator or Excel function, we find that the probability is approximately 0.4871.