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The sampling distribution is the probability distribution of a given statistic based on a random sample. In the context of our exercise, it refers to the distribution of the sample mean heart rates of male runners over several intervals. Although the initial data distribution of heart rates is not normal, the Central Limit Theorem allows us to conclude that the distribution of the sample mean becomes approximately normal when the sample size is large enough.

In our problem, we have 24 intervals, which is sufficiently large for the Central Limit Theorem to apply. Hence, the sampling distribution of the mean number of beats per five seconds can be considered as approximately normal. This assumption simplifies probability calculations and other statistical inferences about the data.

The normal distribution is a continuous probability distribution that is symmetrical around its mean, often described as a bell curve. It's characterized by the parameters mean (\(\mu \)) and standard deviation (\(\sigma \)).

In our scenario, the sample mean heart rates follow a normal distribution because of the Central Limit Theorem. We use this normal approximation with mean \(\mu = 8.8\) beats per five seconds and a standard deviation, or standard error, calculated as \(\sigma = \frac{1.0}{\sqrt{24}} \approx 0.204\). This allows us to easily compute probabilities and make predictions about the runners' heart rates.

Standard deviation measures the amount of variation or dispersion from the average or mean. In the context of this study, the standard deviation of heart rates within five seconds is given as 1.0. This value indicates the variability of single observations within the population.

When examining the sample mean from 24 intervals, the standard deviation is adjusted to what we call the standard error, which is smaller than the population standard deviation. The standard error is calculated by dividing the standard deviation by the square root of the sample size: \(\sigma = \frac{1.0}{\sqrt{24}} \approx 0.204\). This reduced value reflects the expected variability of the sample mean compared to individual observations.

Probability calculation enables us to quantify how likely certain outcomes are given the distribution characteristics. In this problem, we use the properties of the normal distribution to find probabilities.

For example, to find the probability that the mean heart rate is less than 8 beats per five seconds, we compute the Z-score: \(Z = \frac{8 - 8.8}{0.204} = -3.92\). The Z-score helps us transform our random variable into a standard normal distribution for easy lookup in statistical tables or software.

The same approach is used for determining the probability of a heart rate being less than 100 beats per minute. Convert this rate to beats per five seconds (8.33 beats per five seconds) and compute the Z-score: \(Z = \frac{8.33 - 8.8}{0.204} = -2.30\). These Z-scores provide the tools to find the corresponding probabilities with ease, providing insight into the likelihood of observed outcomes.