Sodium in muffins: a sample mean and a skewed population
An industrial bakery fills muffin cups by weight, and sodium content per muffin is right-skewed: the population mean is 210 mg and the population standard deviation is 45 mg, with a few very high values from uneven salt blending. A food inspector plans to take a random sample of 49 muffins and will flag the batch if the sample mean sodium exceeds 220 mg. A second inspector, working at a smaller satellite kitchen, plans to take only 9 muffins and intends to use the same normal calculation with the standard error 45/sqrt(9) = 15 mg. The plant manager also asks whether a single muffin measuring 300 mg should be considered unusual, since that is two standard deviations above the mean. The batch is due for release tomorrow, and the inspectors disagree about whether the sample size is large enough for the normal model.
1. Compute the probability that the sample mean of 49 muffins exceeds 220 mg, using the central limit theorem. State the assumptions and the role of the skewed population.
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For n = 49, the standard error is sigma/sqrt(n) = 45/7 = 6.43 mg. The central limit theorem says the sample mean is approximately normal with mean 210 and SE 6.43 even though individual sodium values are right-skewed, provided the sample is random and n is large enough relative to the skew. Standardize: z = (220 - 210)/6.43 = 1.56. The upper-tail probability is about 0.060, so there is roughly a 6% chance of flagging a batch whose true mean is 210 mg.
Using t with 48 df instead of z gives about 0.063, very similar. The approximation is reasonable at n = 49 for moderate skew, but the population's right tail means the normal approximation for the mean is less exact than for a symmetric population. If the decision is close to the threshold, a bootstrap or a larger sample is more defensible. Random sampling and independence of muffins are also required.
2. The second inspector will use n = 9 and the same normal calculation. Explain why that is unreliable and propose alternatives.
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The standard error 45/sqrt(9) = 15 mg is arithmetically correct, but the central limit theorem justification is not. With n = 9 and a strongly right-skewed population, the sampling distribution of the mean can still be noticeably skewed and non-normal, so a z- or t-based tail probability can be badly wrong. The second inspector should not use the same normal calculation.
Options are to increase the sample to at least 30-50 if the skew is moderate, use a bootstrap confidence interval or permutation test, or model the sodium distribution directly, for example as log-normal, and compute the sampling distribution. A t interval is also sensitive to the heavy right tail at n = 9. If a decision must be made now, report the uncertainty qualitatively or use a conservative nonparametric bound rather than a precise probability such as 0.25.
3. Is a single muffin at 300 mg unusual? Compare the normal-model answer with what the skewness implies, and recommend how to judge it operationally.
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Using a normal model, z = (300 - 210)/45 = 2.0, and the upper-tail probability is 0.0228, so 300 mg would be "unusual" by the 2 SD rule. But the population is right-skewed, meaning high values are more common than a normal model predicts; the true tail probability could be larger than 2.3%. Conversely, a normal model fitted to the bulk could misstate the extreme tail.
Chebyshev's inequality gives a distribution-free bound: at most 1/k^2 = 1/4 of values lie beyond 2 SD, so 300 mg is not impossible, but it is not routine either. A single muffin is a sample of size one; it should trigger a check of the filling process, not a conclusion about the batch mean. To judge unusualness operationally, use the empirical distribution or a log-normal fit and compare the value with process control limits.