Binomial, normal and Poisson models · 二项分布、正态分布与泊松分布模型
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| expected value/ekˈspektɪd ˈvæljuː/ | 期望值 | qī wàng zhí |
What makes a count predictable?
- A quality inspector counts defective items. The number is random, but a model can describe its likely range.
- This lesson studies expected value 期望值: The probability-weighted mean of a random variable.
Choose the mathematical structure
- A binomial model needs fixed n, independent trials, two outcomes and constant p. E(X)=np and Var(X)=np(1-p). For a normal model use z=(x-μ)/σ. State the event before calculating a tail probability.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines expected value? · 以下哪项描述正确定义了期望值？
The probability-weighted mean of a random variable. · 随机变量的概率加权均值。
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
For X binomial(5,0.2), P(X=0)=0.8^5=0.32768, E(X)=1 and Var(X)=0.8. For a normal quantity with μ=100,σ=15, the value 130 has z=2. Poisson is not part of this AQA 7357 lesson.
Binomial, normal and Poisson models · 二项分布、正态分布与泊松分布模型
A binomial model needs fixed n, independent trials, two outcomes and constant p · 二项模型需要固定的n、独立试验、两种结果以及恒定的p。
Compare the model with the worked case and explain one change. · 对比模型与已解案例，并说明其中一处变化。
For X binomial(5,0.2), find P(X=0). · 对于 X~binomial(5,0.2)，求 P(X=0)。
No successes means five failures: 0.8⁵=0.32768. · 零次成功意味着五次失败：0.8⁵=0.32768。
Test a tempting shortcut
- Not every count is binomial: changing p or dependence can invalidate it. For a continuous variable, the probability of one exact value is zero. Continuity correction matters when approximating a discrete distribution by a normal one.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Every count of successes has a binomial distribution regardless of dependence. This claim is false. Explain which definition or assumption it violates.
Find Var(X) for X binomial(5,0.2). · 求X服从二项分布(5,0.2)时的Var(X)。
Binomial variance=np(1-p)=5×0.2×0.8=0.8. · 二项分布方差=np(1-p)=5×0.2×0.8=0.8。
Every count of successes has a binomial distribution regardless of dependence. · 无论是否存在依赖关系，每次成功的计数都服从二项分布。
Not every count is binomial: changing p or dependence can invalidate it. For a continuous variable, the probability of one exact value is zero. Continuity correction matters when approximating a discrete distribution by a normal one. · 并非所有计数都是二项分布：改变p或存在依赖关系可能导致其不再适用。对于连续变量，取某一个精确值的概率为零。当用正态分布近似离散分布时，连续性校正至关重要。
Interpret a new situation
- Write the event as an inequality before using calculator distribution functions. Distinguish P(X<k), P(X≤k) and a tail complement. State assumptions in context.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
For μ=100 and σ=15, find the z-score of 130. · 已知μ=100且σ=15，求130的z分数。
Standardise: z=(130-100)/15=2. · 标准化：z=(130-100)/15=2。
Match each part of a complete solution to its purpose. · 将完整解答的每个部分与其目的相匹配。
An assumption justifies the model; a check tests the result; interpretation connects it to the question. · 假设用于论证模型的合理性；检查用于验证结果；解释用于将其与问题建立联系。
Use this in your course
- 7357 · A-level · N. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The probability-weighted mean of a random variable. Choose the relationship, show the method, check its assumptions and interpret the result.