GB701 Unit 3 Probability Problem Set Example

When the Population Is Known: A Unit 3 Probability Problem Set on How Samples of Accounts Behave

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Purdue University Global

GB701: Business Statistics

Unit 3 Probability Problem Set

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[Date]

The distributor, its accounts and every figure are composite and illustrative; the methods and sources cited are real.

Problem 1: Slow Conversion as a Binomial Event

Question. If 25 accounts are sampled at random, what is the chance that none of them is a slow converter, defined as taking more than 16 days from quote to order?

Rule. Each sampled account either is or is not slow, and the population proportion is known: 100 of 1,240 accounts, p = 0.081. Treating the 25 draws as independent trials gives a binomial distribution. Strictly, sampling without replacement from a finite population follows the hypergeometric distribution, but with 25 drawn from 1,240 the difference is negligible, as Lohr (2019) notes for samples that are a small fraction of their frame.

Working. The probability of no slow account is 0.919, the chance a single draw is not slow, raised to the 25th power, which equals 0.122. The probability of three or more slow accounts in a sample of 25 is 0.328, so even the typical small sample contains only one or two.

Simulation check. Of 10,000 random samples of 25 drawn without replacement, 12.0 percent contained no slow account.

For the sampling plan. A small sample has about a one-in-eight chance of missing the slow group entirely. Any analysis of slow conversion needs either a much larger sample or deliberate stratification.

Problem 2: Conditional Probability from the Ledger

Question. Given that an account is slow, how likely is it to be a special-order account?

Rule. Conditional probability: P(special | slow) = P(special and slow) / P(slow).

Working. Of the 100 slow accounts, 72 are special-order accounts, so P(special | slow) = 0.72. Special-order accounts are 28.1 percent of all accounts. Turning the condition around, 20.6 percent of special-order accounts are slow, against 3.1 percent of stock accounts.

For the sampling plan. Slowness is concentrated in one group. A sample that underrepresents special-order accounts will understate the slow tail, which argues for stratifying by order type.

Problem 3: The Sampling Distribution of the Mean

Question. How much would the mean of a random sample of 216 accounts vary from sample to sample?

Rule. The standard error of a sample mean equals sigma over the square root of n, adjusted downward by a finite population correction whenever the sample is a sizable share of the population.

Working. With a population standard deviation of 5.01 days, the standard error for n = 216 is 5.01 / 14.70 = 0.341 days, and the finite population correction, the square root of 1,024 divided by 1,239, reduces it to 0.310.

Simulation check. The standard deviation of 10,000 simulated sample means was 0.307 days, close to the corrected figure. The means themselves were nearly symmetric, with a skewness of 0.15, even though the population is strongly skewed.

For the sampling plan. The central limit theorem holds well at n = 216. A sample mean will usually fall within about 0.6 days of the population mean.

Problem 4: A Noisier Median

Question. Is the sample median a more or less stable estimate than the sample mean?

Working and simulation. For n = 216, the standard deviation of 10,000 simulated sample medians was 0.319 days, slightly larger than the 0.307 for means. For a skewed distribution the median is the better description of the typical account, but in this population it is a slightly less precise estimator.

For the sampling plan. The median should be reported, but its interval should be estimated by bootstrapping, since no simple formula gives its standard error for a skewed population.

The median describes the typical account better, and the mean estimates its own target more precisely; a good report uses each for what it does best.

What the Five Problems Decide

Taken together, the problems settle three choices for the next unit before any file is opened. The sample must be stratified by order type, because slowness is concentrated among special-order accounts. It must be large enough that the mean is estimated within about half a day, which a sample near 216 achieves. And any interval for the median must come from resampling rather than a formula. Cochran (1977) describes the purpose of such preliminary work as fixing the design before data collection, so that the design cannot be adjusted afterward to suit the results.

Problem 5: Where the Normal Approximation Slips

Question. Does a standard 95 percent t interval for the mean actually capture the population mean 95 percent of the time?

Simulation. For samples of 25, the t interval captured the true mean in 93.7 percent of 10,000 samples, falling short because the skewed population produces skewed sample means at that size; their skewness was 0.33. For samples of 216, coverage was 96.9 percent, above the nominal level, because the interval ignores the finite population correction and is therefore slightly too wide.

For the sampling plan. At n = 216 the t interval is safe and slightly conservative. At n = 25 it would understate uncertainty. Efron and Tibshirani (1993) describe the bootstrap as a way to check such approximations when a population's shape is unknown, which will matter once the analysis moves from the known frame to a coded sample whose variables cannot be checked against a population.

References

Cochran, W. G. (1977). Sampling techniques (3rd ed.). Wiley.

Efron, B., & Tibshirani, R. J. (1993). An introduction to the bootstrap. Chapman & Hall.

Lohr, S. L. (2019). Sampling: Design and analysis (2nd ed.). CRC Press.