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2.9 Mean and Spread of a Discrete Distribution — Answers and Rubrics

These solutions explain the new practice tasks in this lesson. For simulation parts, a valid recorded run may differ from another learner’s run; the rubric checks the model, recorded work and interpretation.

Original generated practice for the Fall 2026 course. Return to the lesson and independent checks.

Calculate Daily Booking Parameters: worked answer

Known model for daily bookings at one rental kiosk
Bookings X0124
Probability.10.20.40.30

This supplied fixed model governs the population of days under a specified operating policy. X is the number of bookings on one day; it is not a sample summary.

Part a

Identify the possible values and probabilities, explain why the distribution mean is a fixed parameter, and calculate E(X)=μX using a complete weighted sum.

The values are 0,1,2,4 with the stated probabilities. Their mean is a single fixed characteristic of this probability model, unlike an individual day or a varying sample average. E(X)=μX=0(.10)+1(.20)+2(.40)+4(.30)=2.20E(X)=\mu_X=0(.10)+1(.20)+2(.40)+4(.30)=2.20 bookings per model-day.

Part a scoring criteria
Evidence for creditPoints
Identifies a fixed distribution parameter and variable days/sample averages.1
Shows the complete weighted sum and obtains 2.20 bookings.1

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Part b

Calculate V(X) and SD(X) with the weighted squared-deviation formula. Show each contribution and distinguish notation and units.

The contributions are (0−2.2)2(.10)=.484(0-2.2)^2(.10)=.484, (1−2.2)2(.20)=.288(1-2.2)^2(.20)=.288, (2−2.2)2(.40)=.016(2-2.2)^2(.40)=.016, and (4−2.2)2(.30)=.972(4-2.2)^2(.30)=.972. Their sum is V(X)=σX2=1.76V(X)=\sigma_X^2=1.76 squared bookings; SD(X)=σX=1.76≈1.327SD(X)=\sigma_X=\sqrt{1.76}\approx1.327 bookings. These are fixed model parameters.

Part b scoring criteria
Evidence for creditPoints
Shows all four correct weighted squared deviations.1
Obtains variance 1.76 and SD about 1.327.1
Distinguishes variance/SD notation and squared/original units.1

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Interpret a Delivery Distribution: worked answer

Known fixed model for one day of deliveries to a neighborhood desk
Kits delivered X025
Probability.20.50.30

For this defined population of operating days, the model mean is 2.5 kits and its standard deviation is √3.25≈1.803 kits.

Part a

Interpret the mean and standard deviation in the defined population, including the long-run and typical-deviation meanings. Explain why no day has to deliver exactly 2.5 kits and why SD does not mean the day-to-day change.

Over many operating days under this fixed neighborhood-desk model, the average number delivered approaches 2.5 kits per day. A daily delivery count typically differs about 1.803 kits from that mean. Values are countable (0,2,5); an average can be 2.5 even though no day has that value. SD summarizes distances from the population mean, not differences between consecutive days.

Part a scoring criteria
Evidence for creditPoints
Interprets 2.5 as the long-run daily average in the supplied population.1
Interprets 1.803 as typical deviation of daily counts from 2.5.1
Explains nonattainable average and rejects day-to-day-change meaning.1

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