The Owner Of A Fish Market Has An Assistant Who Has Determin

The owner of a fish market has an assistant who has determined that

The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with a mean of 3.2 pounds and a standard deviation of 0.8 pound. The question asks: What percentage of samples of 4 fish will have sample means between 3.0 and 4.0 pounds?

To solve this problem, we need to understand the concept of sampling distribution of the sample mean. Since the weights of individual catfish are normally distributed, the distribution of the sample mean for a sample size of 4 fish will also be normally distributed, with the same mean as the population mean, but with a reduced standard deviation (standard error).

The mean of the sampling distribution (μₓ̄) is equal to the population mean, which is 3.2 pounds.

The standard deviation of the sampling distribution (standard error, SE) is calculated as:

SE = σ / √n = 0.8 / √4 = 0.8 / 2 = 0.4 pounds.

Next, we determine the probability that the sample mean falls between 3.0 and 4.0 pounds.

We convert these to their corresponding z-scores:

Z = (X̄ – μ) / SE

For X̄ = 3.0 pounds:

Z₁ = (3.0 – 3.2) / 0.4 = -0.2 / 0.4 = -0.5

For X̄ = 4.0 pounds:

Z₂ = (4.0 – 3.2) / 0.4 = 0.8 / 0.4 = 2.0

Using standard normal distribution tables or a calculator, find the probabilities:

P(Z

P(Z

Therefore, the probability that the sample mean is between 3.0 and 4.0 pounds is:

P(−0.5

Expressed as a percentage, this is approximately 66.87%.

Looking at the options provided:

a) 84%

b) 67%

c) 29%

d) 16%

The closest percentage is 67%.

Therefore, approximately 67% of samples of 4 fish will have sample means between 3.0 and 4.0 pounds.

Paper For Above instruction

In statistical analysis, understanding the behavior of sample means in relation to the population parameters is crucial, especially when dealing with normally distributed data. The problem presented involves determining the proportion of sample means falling within a specific interval, given the population mean and standard deviation, for a sample size of four fish. This relates directly to the concept of the sampling distribution of the mean, which underpins many inferential statistics techniques.

The population parameters specify that the weights of catfish in the market are normally distributed with a mean of 3.2 pounds and a standard deviation of 0.8 pound. When taking samples of four fish, the distribution of the sample mean also follows a normal distribution by the properties of normality and the Central Limit Theorem. This distribution has the same mean as the population, 3.2 pounds, but with a reduced standard deviation, known as the standard error, calculated by dividing the population standard deviation by the square root of the sample size.

For the given sample size of four, the standard error is:

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{0.8}{\sqrt{4}} = 0.4 \]

Thus, the sampling distribution of the mean is normally distributed with a mean of 3.2 pounds and a standard deviation of 0.4 pounds.

To find the proportion of samples with means between 3.0 and 4.0 pounds, we convert these bounds into z-scores:

- For 3.0 pounds:

\[ Z = \frac{X̄ - \mu}{SE} = \frac{3.0 - 3.2}{0.4} = -0.5 \]

- For 4.0 pounds:

\[ Z = \frac{4.0 - 3.2}{0.4} = 2.0 \]

Using the standard normal distribution table or a calculator, the corresponding probabilities are:

- \( P(Z

- \( P(Z

Subtracting these gives the probability that the sample mean falls between 3.0 and 4.0 pounds:

\[ P(-0.5

Expressed as a percentage, this approximation indicates that about 66.87%, or roughly 67%, of such samples would have mean weights between these bounds.

This analysis is significant in quality control processes, where understanding the likelihood of sample means falling within specified ranges helps in making informed decisions about process consistency. The calculation hinges integral on the properties of the normal distribution, the Central Limit Theorem, and the concept of standard error, which simplifies the understanding of how sample means behave relative to the population parameters.

In conclusion, given the population weight of catfish and the sample size of four, approximately 67% of samples will have means lying between 3.0 and 4.0 pounds. This result not only serves as an example of applying normal distribution principles but also demonstrates the practical application of statistical inference in real-world scenarios such as food quality inspection.

References

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