Tuatara are rare, medium-sized reptiles found only in New Zealand. Assume female lengths are normally distributed.The length of each tuatara in a sample of 10 female tuatara is measured giving an average length of 48.5cm with a standard deviation of 2.9cm.Suppose we repeatedly took random samples of 10 female tuatara. What percentage of these samples would we expect to have a sample mean weight that is less than 50.34cm?(You should use the empirical rules to estimate this percentage.)Select one:a.2.5%b.100%c.95%d.97.5%
Question
Tuatara are rare, medium-sized reptiles found only in New Zealand. Assume female lengths are normally distributed.The length of each tuatara in a sample of 10 female tuatara is measured giving an average length of 48.5cm with a standard deviation of 2.9cm.Suppose we repeatedly took random samples of 10 female tuatara. What percentage of these samples would we expect to have a sample mean weight that is less than 50.34cm?(You should use the empirical rules to estimate this percentage.)Select one:a.2.5%b.100%c.95%d.97.5%
Solution
The empirical rule, also known as the 68-95-99.7 rule, states that for a normal distribution:
- About 68% of the data falls within one standard deviation of the mean.
- About 95% falls within two standard deviations.
- About 99.7% falls within three standard deviations.
First, we need to calculate the standard error (SE) of the sample mean. The formula for the standard error is:
SE = σ / √n
where: SE = standard error σ = standard deviation n = sample size
In this case, the standard deviation (σ) is 2.9 cm and the sample size (n) is 10.
Substituting these values into the formula gives:
SE = 2.9 / √10 ≈ 0.917 cm
Next, we calculate the z-score, which is the number of standard errors a particular value is from the mean. The formula for the z-score is:
Z = (X - μ) / SE
where: Z = z-score X = value of interest μ = mean SE = standard error
In this case, the value of interest (X) is 50.34 cm, the mean (μ) is 48.5 cm, and the standard error (SE) is 0.917 cm.
Substituting these values into the formula gives:
Z = (50.34 - 48.5) / 0.917 ≈ 2.01
A z-score of 2.01 is just over two standard deviations from the mean. According to the empirical rule, about 95% of the data falls within two standard deviations of the mean. This means that about 5% of the data falls outside two standard deviations. Since the z-score is positive, we are interested in the percentage of data that is greater than two standard deviations above the mean, which is 2.5% (half of 5%).
Therefore, we would expect about 2.5% of the samples to have a sample mean weight that is less than 50.34 cm.
So, the correct answer is:
a. 2.5%
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