The random variable WWW represents the initial nutrient concentration, in mg/L, in a series of botanical samples. The probability distribution for WWW is given in the following table:
| www | 1 | 3 | 6 | 8 |
|---|---|---|---|---|
| P(W=w)P(W=w)P(W=w) | 0.45 | 0.2 | 0.2 | 0.15 |
Show that E(W)=3.45E(W) = 3.45E(W)=3.45.
Find Var(W)Var(W)Var(W).
The random variable SSS represents the soil porosity index of the sample's medium. The probability distribution for SSS is given in the following table, where kkk is a constant:
| sss | 2 | 4 | 5 | kkk |
|---|---|---|---|---|
| P(S=s)P(S=s)P(S=s) | 0.25 | 0.25 | 0.25 | 0.25 |
Name the probability distribution of SSS.
Given that E(S)=E(W)E(S) = E(W)E(S)=E(W), find the value of kkk.
The growth of a seedling, GGG mm, is modelled by the normal distribution G∼N(μ,σ2)G \sim N(\mu, \sigma^2)G∼N(μ,σ2). Researchers Alice and Bob each select a nutrient concentration for μ\muμ and a porosity index for σ\sigmaσ by sampling from the distributions of WWW and SSS respectively. A sample is considered 'successful' if its growth exceeds 5 mm. The researcher whose parameters result in a higher probability of success, P(G>5)P(G > 5)P(G>5), wins.
Alice obtained w=6w = 6w=6 and s=2s = 2s=2. Bob obtained s=5s = 5s=5. Determine the probability that Bob wins.
Find the largest probability of success, P(G>5)P(G > 5)P(G>5), achievable in this experiment.
Assuming that the selections of W and S are independent, find the probability of a researcher achieving this maximum probability of success, P(G > 5).
616 exam-style questions on Edexcel A Level Maths The Normal Distribution, covering 3.1 The Normal Distribution, 3.2 Finding Probabilities for Normal Distributions, 3.3 The Inverse Normal Distribution Function, 3.4 The Standard Normal Distribution, 3.5 Finding the mean and standard deviation, 3.6 Approximating a Binomial Distribution, and 3.7 Hypothesis Testing with the Normal Distribution. Each one has a worked solution and a mark scheme showing where the marks go.