Assuming Forestry Commission data, a random sample of 400 UK hectares has an ~11% chance of showing "sustainably managed" proportion exceeding 47%, thanks to the Central Limit Theorem and z-score analysis.
Option D is correct.
Here's the analysis of the situation and the approximate probability:
Information:
Total hectares: 3.16 million
Sustainable hectares (reported): 43%
Sample size: 400 hectares
Sample percentage with certification: 47%
Assumptions:
Simple random sampling ensures every hectare has an equal chance of being included.
The Forestry Commission report is accurate.
Probability Calculation:
We need to find the probability (P) that more than 47% of the sample (400 hectares) have the "sustainably managed" certification. This can be approximated using a normal distribution (also called the Central Limit Theorem) because:
Sample size (400) is sufficiently large.
We have no evidence suggesting significant non-normality in the data.
Therefore, we can estimate the probability using the:
Mean (µ) of the "sustainably managed" proportion: 0.43 (from the overall population)
Standard deviation (σ) of the proportion: calculated as √(p*(1-p)/n) ≈ √(0.43*0.57/400) ≈ 0.035
Now, we can calculate the probability (P) of finding a sample proportion (ṗ) greater than 0.47:
P(ṗ > 0.47) ≈ 1 - P(ṗ ≤ 0.47)
We can use a z-score table or calculator to find P(ṗ ≤ 0.47). The z-score for 0.47 with the estimated mean and standard deviation is approximately 1.14.
Looking up the z-score table, we find P(ṗ ≤ 0.47) ≈ 0.8710. Finally, subtracting this from 1:
P(ṗ > 0.47) ≈ 1 - 0.8710 ≈ 0.1290
Approximate Answer:
D. P(ṗ>0.47) ≈ 0.11 (rounded to 2 decimal places)
Therefore, the approximate probability of finding a sample proportion exceeding 47% "sustainably managed" hectares is around 11%.