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4. The department of fish and wildlife want to estimate the size of alligator population in

Louisiana. The department issued 937 tags to hunters in 2018. The average length of the
alligators was 11.2 ft with a standard deviation of 2.06 feet. What is the mean and
standard deviation of the population?

User Westporch
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We can use the information provided to estimate the mean and standard deviation of the alligator population in Louisiana.

Let's start by using the formula for the standard error of the mean:

SE = s / sqrt(n)

where s is the sample standard deviation, n is the sample size, and SE is the standard error of the mean. The standard error of the mean estimates the standard deviation of the population based on the sample data.

Substituting the values given, we get:

SE = 2.06 / sqrt(937)
SE ≈ 0.067

Next, we can use the formula for the confidence interval of the mean:

CI = X ± (Z * SE)

where X is the sample mean, Z is the z-score associated with the desired level of confidence, and SE is the standard error of the mean. The confidence interval estimates the range within which the population mean is likely to fall based on the sample data.

Assuming a 95% level of confidence (which corresponds to a z-score of 1.96), we get:

CI = 11.2 ± (1.96 * 0.067)
CI ≈ (11.07, 11.33)

Therefore, we can estimate that the mean length of the alligator population in Louisiana is between 11.07 and 11.33 feet.

To estimate the standard deviation of the population, we can use the formula:

s = sqrt(((n-1) * s^2) / n)

where s is the sample standard deviation, n is the sample size, and s is the population standard deviation. This formula estimates the population standard deviation based on the sample data.

Substituting the values given, we get:

s = sqrt(((937-1) * 2.06^2) / 937)
s ≈ 2.03

Therefore, we can estimate that the standard deviation of the alligator population in Louisiana is approximately 2.03 feet.
User Nanachan
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