Let's go through the questions step by step:

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Part (b): Probability that at most 4 seeds grow (out of 11)

This is a binomial probability problem where the total number of trials is $n = 11$ and we need to find the probability of at most 4 successes (i.e., plants grow).

The general formula for binomial probability is:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

Where:

• $n = 11$ (number of seeds planted),
• $p$ is the probability that a seed grows (which we assume is the same as in part (a), but it needs to be specified),
• $X = k$ is the number of seeds that grow.

For "at most 4 grow", we need to compute the cumulative probability:

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

The specific probability $p$ of a seed growing is crucial here. If you provide that, I can compute the exact probability for you.

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Part (c): Normal approximation to the binomial

In large binomial distributions (like $n = 580$), we can approximate the binomial distribution using the normal distribution. The key condition for this approximation is that both $np$ and $n(1-p)$ should be sufficiently large (greater than 5). For this, we need the probability $p$ of a seed growing to calculate both successes and failures.

The normal approximation would be based on:

• Mean $\mu = np$
• Standard deviation $\sigma = \sqrt{np(1-p)}$

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Part (d): Mean and standard deviation (for 580 seeds)

We can compute the mean and standard deviation for the number of plants expected to grow as follows:

• Mean: $\mu = np$ Where $n = 580$ and $p$ is the probability of success (growing into a healthy plant).

• Standard deviation: $\sigma = \sqrt{np(1-p)}$

I will need the probability $p$ to provide specific values.

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Part (e): Probability that fewer than 372 grow (with normal approximation)

Using the normal approximation to the binomial distribution, we can calculate the probability that fewer than 372 plants grow. The formula is:

$P(X < 372) \approx P\left(Z < \frac{372 - \mu}{\sigma}\right)$

Where $Z$ is the standard normal variable, and $\mu$ and $\sigma$ are the mean and standard deviation calculated earlier. With the correct value of $p$, I can solve this using the z-score table.

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Please provide the probability $p$ that a seed will grow, and I'll compute the exact values for parts (b), (d), and (e).

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Relative questions:

1. How does the binomial distribution differ from the normal distribution?
2. Why is the normal approximation used for large sample sizes?
3. What are the conditions for using the normal approximation to the binomial distribution?
4. How would the probability change if more seeds are planted (e.g., 1000 instead of 580)?
5. Can you explain how the z-score helps in finding probabilities for normal distributions?

Tip:

When working with binomial distributions, always check if the normal approximation applies by ensuring $np \geq 5$ and $n(1-p) \geq 5$.