The St. Joe Company grows pine trees and the average annual increase in tree diameter is 3.1 inches (µ) with a population standard deviation of 0.5 inches. A random sample of n = 50 trees is collected. What is the probability of the sample mean being less the 2.9 inches?

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BlackTom AI · Accepted Answer

We start by identifying the distribution of the sample mean. The population mean is µ = 3.1 inches and the population standard deviation is σ = 0.5 inches. With n = 50, the sampling distribution of the sample mean X̄ is approximately Normal with mean µ_X̄ = µ = 3.1 and standard deviation SE = σ/√n = 0.5/√50 ≈ 0.07071 inches.

Now we want P(X̄ < 2.9). Convert to a Z score: Z = (X̄ − µ_X̄) / SE = (2.9 − 3.1) / 0.07071 ≈ −0.2 / 0.07071 ≈ −2.83.

Option-by-option analysis:
- Option 0.4977: This would correspond to a Z value near 0 (since P(Z < 0) = 0.5). Our computed Z is about −2.83, which is far from 0 and corresponds to a much smaller left-tail probability. Therefore this option overestimates the likelihood and is inconsistent with the calculated Z. 
- Option 0.0023: A Z of −2.83 sits well in the left tail of the standard normal distribution. The standard normal CDF Φ(−2.83) is approximately 0.0023, which matches the computed tail probability for P(X̄ < 2.9). This aligns with the expected very small probability of observing a mean this low purely by sampling variability. 
- Option 0.9977: This is the complement of the previous tail and would correspond to P(Z < −2.83) being near 0.9977, which would imply almost certain occurrence of such a low mean, contradicting the actual left-tail calculation. This option incorrectly interprets the direction and magnitude of the tail probability.
- Option 0.9954: This value is also a high left-tail-like probability but still inconsistent with the Z of −2.83. It would imply a much larger probability than what the standard normal distribution yields for such a negative Z, so it does not fit the calculated tail area.

In summary, the correct probability is the one that matches the Z ≈ −2.83 in the standard normal distribution, which corresponds to a small left-tail probability around 0.0023. The other options either approximate near the center (around 0.5), or misrepresent the tail probability for the given Z value, leading to implausible results given the calculation.

OSC 2030-009 Fall 2025 Exam 2

The St. Joe Company grows pine trees and the average annual increase in tree diameter is 3.1 inches (µ) with a population standard deviation of 0.5 inches. A random sample of n = 50 trees is collected. What is the probability of the sample mean being less the 2.9 inches?

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