Sampling Dynamics Worksheet
Sampling Dynamics
Statistical Analysis & Probability Worksheet
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Case Study: CHAOS Strategy
Is there a different side to random sampling? Can sampling be used as a weapon? The acronym for Create Havoc Around Our System is CHAOS. The Association of Flight Attendants (AFA) used this as a negotiation tool against Alaska Airlines in 1994.
CHAOS involves a small sample of random strikes—a few flights at a time—instead of a mass walkout. This makes the entire schedule unreliable, which management cannot tolerate. In 1986, TWA flight attendants were permanently replaced after a mass walkout. Using CHAOS, only a few jobs are at risk, and random sampling becomes a powerful strategic weapon.
Statistical Problems
General: Suppose that \(x\) has a distribution with \(\mu = 15\) and \(\sigma = 14\).
(a) For \(n = 49\), find \(\mu_{\bar{x}}, \sigma_{\bar{x}}\), and \(P(15 \le \bar{x} \le 17)\).
(b) For \(n = 64\), find \(\mu_{\bar{x}}, \sigma_{\bar{x}}\), and \(P(15 \le \bar{x} \le 17)\).
(c) Why is the probability in (b) higher than (a)? (Hint: Consider standard deviations.)
General: Suppose that \(x\) has a distribution with \(\mu = 100\) and \(\sigma = 48\).
(a) For \(n = 81\), find \(\mu_{\bar{x}}, \sigma_{\bar{x}}\), and \(P(92 \le \bar{x} \le 100)\).
(b) For \(n = 121\), find \(\mu_{\bar{x}}, \sigma_{\bar{x}}\), and \(P(92 \le \bar{x} \le 100)\).
(c) Comment on the differences in probabilities and why you expect them.
General: Suppose that \(x\) has a distribution with \(\mu = 25\) and \(\sigma = 3.5\).
(a) If random samples of size \(n = 9\) are selected, can we say anything about the \(\bar{x}\) distribution of sample means?
(b) If the original \(x\) distribution is normal, find \(P(23 \le \bar{x} \le 26)\) for samples of size \(n = 9\).
General: Suppose that \(x\) has a distribution with \(\mu = 72\) and \(\sigma = 8\).
(a) If random samples of size \(n = 16\) are selected, can we say anything about the \(\bar{x}\) distribution?
(b) If the original \(x\) distribution is normal, find \(P(68 \le \bar{x} \le 73)\) for \(n = 16\).
Case: Coal Loader Efficiency
A loader puts coal into hopper cars with weights normally distributed (\(\mu = 75\) tons, \(\sigma = 0.8\) ton).
(a) Find the probability that one car chosen at random has less than 74.5 tons.
(b) Find the probability that the mean weight of \(n = 20\) cars is less than 74.5 tons.
Sampling Dynamics Answer Key
Answer Key
Sampling Dynamics: Solutions & Explanations
TEACHER RESOURCE
Problem 1 (\(\mu=15, \sigma=14\))
(a) \(n=49\): \(\mu_{\bar{x}} = 15\); \(\sigma_{\bar{x}} = 14/\sqrt{49} = 2\); \(P(15 \le \bar{x} \le 17) = P(0 \le z \le 1) = \mathbf{0.3413}\)
(b) \(n=64\): \(\mu_{\bar{x}} = 15\); \(\sigma_{\bar{x}} = 14/8 = 1.75\); \(P(15 \le \bar{x} \le 17) = P(0 \le z \le 1.14) \approx \mathbf{0.3729}\)
(c) Reason: Larger sample size decreases the standard error, making the distribution tighter. More probability is concentrated near the mean.
Problem 2 (\(\mu=100, \sigma=48\))
(a) \(n=81\): \(\mu_{\bar{x}} = 100\); \(\sigma_{\bar{x}} = 48/9 = 5.33\); \(P(92 \le \bar{x} \le 100) = P(-1.5 \le z \le 0) = \mathbf{0.4332}\)
(b) \(n=121\): \(\mu_{\bar{x}} = 100\); \(\sigma_{\bar{x}} = 48/11 = 4.36\); \(P(92 \le \bar{x} \le 100) = P(-1.83 \le z \le 0) \approx \mathbf{0.4664}\)
(c) Reason: Increased \(n\) results in a smaller standard error, leading to a higher concentration of area under the curve near the population mean.
Problem 3 (\(\mu=25, \sigma=3.5\))
(a) Sample size \(n=9\) is too small (\(<30\)). Without knowing population shape, we cannot assume normality for \(\bar{x}\).
(b) Normal Population: \(\sigma_{\bar{x}} \approx 1.17\). \(P(23 \le \bar{x} \le 26) = P(-1.71 \le z \le 0.85) \approx \mathbf{0.7587}\)
Problem 4 (\(\mu=72, \sigma=8\))
(a) CLT does not apply (\(n=16 < 30\)). Shape of \(\bar{x}\) distribution is unknown.
(b) Normal Population: \(\sigma_{\bar{x}} = 2\). \(P(68 \le \bar{x} \le 73) = P(-2 \le z \le 0.5) \approx \mathbf{0.6687}\)
Problem 5 (\(\mu=75, \sigma=0.8\))
(a) Single Car: \(P(x < 74.5) = P(z < -0.625) \approx \mathbf{0.2660}\)
(b) Sample \(n=20\): \(\sigma_{\bar{x}} \approx 0.1789\). \(P(\bar{x} < 74.5) = P(z < -2.79) \approx \mathbf{0.0026}\)
