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The United Kingdom Forestry Commission reports that 43%, of the 3.16 million hectares of woodland area in the United Kingdom had certification identifying them as "sustainably managed" in 2016. Suppose an employee took a simple random sample of 400 of the hectares and saw that the records showed that 47% of the sampled hectares had that certification in 2016. Assuming that the Forestry Commission's report is accurate, what is the approximate probability that more than 47% of the sample would have had the certification in 2016?

A. P(ṗ>0.47) ≈ 0,05
B. P(ṗ>0.47) ≈ 0,07
C. P(ṗ>0.47) ≈ 0,09
D. P(ṗ>0.47) ≈ 0,11

User Rdelrossi
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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%.

User Brotcrunsher
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