2.4 Probability Distributions

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Question 35

The random variable $W$ represents the initial nutrient concentration, in mg/L, in a series of botanical samples. The probability distribution for $W$ is given in the following table:

$w$	1	3	6	8
$P (W = w)$	0.45	0.2	0.2	0.15

Show that $E (W) = 3.45$ .

[2]

Find $V a r (W)$ .

[3]

The random variable $S$ represents the soil porosity index of the sample's medium. The probability distribution for $S$ is given in the following table, where $k$ is a constant:

$s$	2	4	5	$k$
$P (S = s)$	0.25	0.25	0.25	0.25

Name the probability distribution of $S$ .

[1]

Given that $E (S) = E (W)$ , find the value of $k$ .

[2]

The growth of a seedling, $G$ mm, is modelled by the normal distribution $\sim N(\mu, \sigma^2)$ . Researchers Alice and Bob each select a nutrient concentration for $μ\mu$ and a porosity index for $σ\sigma$ by sampling from the distributions of $W$ and $S$ respectively. A sample is considered 'successful' if its growth exceeds 5 mm. The researcher whose parameters result in a higher probability of success, $P (G > 5)$ , wins.

Alice obtained $w = 6$ and $s = 2$ . Bob obtained $s = 5$ . Determine the probability that Bob wins.

[3]

Find the largest probability of success, $P (G > 5)$ , achievable in this experiment.

[2]

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).

[1]

w

P (W = w)

0.45

0.2

0.15

s

k

P (S = s)

0.25

2.4 Probability Distributions

Question 35

Markscheme

2.4 Probability Distributions

Question 35

Markscheme

2.4 Probability Distributions

Question 35

2.4 Probability Distributions Questions

2.4 Probability Distributions

Question 35

2.4 Probability Distributions Questions