Learn with Amar
Teaching video
A teaching recording for this chapter has not been published yet. Start with the worked example below and explore the related animations where available.
Browse grades and teaching videos01 · Read and understand
What you will learn
- Validate a discrete probability distribution.
- Compute expected values from weights.
- Distinguish a model average from a possible daily outcome.
Before you start
Decimals, weighted averages, and a linear cost expression.
Keep paper nearby. Read the question once for the context, then again to identify what is known and what you need to find.
Start with a question
Daily spare-part demand D is 0, 1, or 2 units with probabilities 0.5, 0.3, and 0.2. A fictional daily handling cost is C = 4 + 3D dollars. Find expected demand and expected cost.
Why this math matters
Individual days have whole-number demand, but a planning average can be fractional. Expected value weights each possible outcome by its probability, describing a model average across repeated comparable situations. It can help compare plans while leaving room for the fact that any one day's demand may differ substantially from that average.
Set up the model
A useful answer starts with clear assumptions:
- These three demand values are exhaustive and their probabilities remain stable.
- The cost includes a four-dollar fixed charge plus three dollars per demanded unit.
- The calculation is a fictional planning model, with no claim about a real business's profit.
02 · Work through the example
Follow the reasoning, one step at a time.
Try to predict the next step before reading it. After each calculation, explain why the operation makes sense and how it helps answer the original question.
Work through it with Amar
See this example unfold.
The complete worked example, one idea at a time.
How can expected demand be less than one item?
PausedQuestion: Start with the question. Paused.
Question
Start with the question
Daily spare-part demand D is 0, 1, or 2 units with probabilities 0.5, 0.3, and 0.2. A fictional daily handling cost is C = 4 + 3D dollars. Find expected demand and expected cost.
Before you calculate
Read what is known and what you need to find. Make a prediction before moving to the first calculation.
Starts paused. Play advances through the full text at a reading pace; pause whenever you need more time. Previous, Next, and the phase buttons let you set your own pace. Playback pauses when this walkthrough leaves the screen or you switch tabs.
Your device’s reduced-motion setting keeps each phase still. Manual controls remain available. The full written solution stays below.
Check and average the demand
0.5 + 0.3 + 0.2 = 1; E[D] = 0(0.5) + 1(0.3) + 2(0.2) = 0.7
The probabilities are nonnegative and total one. The expected demand is a weighted average, not a demand value that must occur.
Translate each outcome into cost
C values = 4, 7, 10; E[C] = 4(0.5) + 7(0.3) + 10(0.2) = $6.10
Weight the possible daily costs with the same probabilities. Linearity also gives 4 + 3E[D] = 6.10.
Measure the demand's spread
E[D²] = 0 + 0.3 + 0.8 = 1.1; Var(D) = 1.1 − 0.7² = 0.61
The variance indicates uncertainty around the expectation. An average alone cannot tell us how many spare parts are needed on every day.
The result
Expected demand is 0.7 units per day and expected daily handling cost is $6.10.
Ten comparable days have expected total demand seven by linearity of expectation. Independence is not needed for that sum of expectations, but it would matter for many calculations about the total's variance or probability distribution.
Common mistakes to catch
- Rounding expected demand to one before calculating cost changes the answer.
- Budgeting exactly the expectation does not guarantee that every individual day's cost is covered.
03 · Practice independently
Try it before revealing the answer.
Use paper or a calculator as needed. Write your units and reasoning, then open the hint or explanation to check your approach.
Practice 1
If the fixed cost becomes $5 and the per-unit cost stays $3, what is expected cost?
Show a hint
Use 5 + 3E[D].
Reveal answer and explanation
$7.10
5 + 3(0.7) = 7.10 dollars.
Practice 2
What is the probability that the original daily cost exceeds $7?
Show a hint
Identify which demand values make 4 + 3D > 7.
Reveal answer and explanation
0.2
Only D = 2 gives a cost above seven dollars, and that outcome has probability 0.2.
Take the idea with you
Use an expected value for average planning, then separately assess variability and the consequences of unusually high outcomes.
04 · Reflect and continue
Can you explain it in your own words?
Before moving on, explain the main idea without looking at the worked example. Try both practice questions, check your reasoning, and name one mistake you now know how to avoid. Return to a step if you still need support.
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