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In a national forest, the trees are infested with a particular species of

parasite, having a mean of 7.14 parasites per tree and a standard

deviation of 2.61 parasites per tree. A random sample of 48 trees is studied

thoroughly.

Approximate the probability that the mean number of parasites in the

sample is at least 7.5. ____

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Approximate the probability that the mean number of parasites in the

sample is no more than 6.5.

Verified Answer :

To solve this problem, we can use the Central Limit Theorem since the sample size is large (n = 48).

For the first part, to find the probability that the mean number of parasites in the sample is at least 7.5, we need to standardize the sample mean using the z-score formula:

\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]

where:
- \(\bar{x}\) is the sample mean,
- \(\mu\) is the population mean,
- \(\sigma\) is the population standard deviation, and
- \(n\) is the sample size.

Given:
- \(\mu = 7.14\),
- \(\sigma = 2.61\),
- \(n = 48\), and
- we want to find \(P(\bar{x} \geq 7.5)\).

First, let's calculate the z-score:

\[ z = \frac{7.5 - 7.14}{\frac{2.61}{\sqrt{48}}} \]

\[ z = \frac{0.36}{\frac{2.61}{\sqrt{48}}} \]

\[ z \approx \frac{0.36}{0.377} \]

\[ z \approx 0.953 \]

Now, we find the probability using a standard normal distribution table or a calculator. The probability that the z-score is greater than 0.953 is approximately 0.1717.

For the second part, to find the probability that the mean number of parasites in the sample is no more than 6.5, we follow a similar process.

\[ z = \frac{6.5 - 7.14}{\frac{2.61}{\sqrt{48}}} \]

\[ z = \frac{-0.64}{\frac{2.61}{\sqrt{48}}} \]

\[ z \approx \frac{-0.64}{0.377} \]

\[ z \approx -1.697 \]

Again, we find the probability using a standard normal distribution table or a calculator. The probability that the z-score is less than -1.697 is approximately 0.0446.

So, the approximate probabilities are:
- \( P(\bar{x} \geq 7.5) \approx 0.1717 \)
- \( P(\bar{x} \leq 6.5) \approx 0.0446 \)

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