A so-called "murder hornet" has an average body length of 45 mm with a standard deviation of 3 mm. In a random sample of 40 murder hornets, what is the probability that the sample mean of the body length of the sample is between 44.5 and 45.5? 0.854 0.146 0.708 0.932

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### Probability Question on "Murder Hornet" Body Lengths

**Question:**

A so-called "murder hornet" has an average body length of 45 mm with a standard deviation of 3 mm. In a random sample of 40 murder hornets, what is the probability that the sample mean of the body length of the sample is between 44.5 and 45.5?

**Choices:**

1. 0.854
2. 0.146
3. 0.708
4. 0.932

When solving this problem, you would typically use concepts from statistics, particularly those involving the sampling distribution of the sample mean. You can use the z-score formula for this, given that the sample size implies the use of the Central Limit Theorem:

\[ Z = \frac{(\bar{X} - \mu)}{(\sigma / \sqrt{n})} \]

Where:
- \( \bar{X} \) is the sample mean
- \( \mu \) is the population mean
- \( \sigma \) is the standard deviation
- \( n \) is the sample size

In this particular case:
- \( \mu = 45 \, \text{mm} \)
- \( \sigma = 3 \, \text{mm} \)
- \( n = 40 \)

After computing the z-scores corresponding to 44.5 mm and 45.5 mm, you can find the probability by looking up these z-scores in a standard normal distribution table or using a statistical software tool.
Transcribed Image Text:### Probability Question on "Murder Hornet" Body Lengths **Question:** A so-called "murder hornet" has an average body length of 45 mm with a standard deviation of 3 mm. In a random sample of 40 murder hornets, what is the probability that the sample mean of the body length of the sample is between 44.5 and 45.5? **Choices:** 1. 0.854 2. 0.146 3. 0.708 4. 0.932 When solving this problem, you would typically use concepts from statistics, particularly those involving the sampling distribution of the sample mean. You can use the z-score formula for this, given that the sample size implies the use of the Central Limit Theorem: \[ Z = \frac{(\bar{X} - \mu)}{(\sigma / \sqrt{n})} \] Where: - \( \bar{X} \) is the sample mean - \( \mu \) is the population mean - \( \sigma \) is the standard deviation - \( n \) is the sample size In this particular case: - \( \mu = 45 \, \text{mm} \) - \( \sigma = 3 \, \text{mm} \) - \( n = 40 \) After computing the z-scores corresponding to 44.5 mm and 45.5 mm, you can find the probability by looking up these z-scores in a standard normal distribution table or using a statistical software tool.
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