Powelliphanta, New Zealand’s native snails, are among the largest snails in the world. Their weight has a normal distribution with mean of 85 grams and standard deviation of 6.4 grams. Suppose we repeatedly took random samples of 10 of these snails.The percentage of samples that have a sample mean weight less than 81grams is closest to:
Question
Powelliphanta, New Zealand’s native snails, are among the largest snails in the world. Their weight has a normal distribution with mean of 85 grams and standard deviation of 6.4 grams. Suppose we repeatedly took random samples of 10 of these snails.The percentage of samples that have a sample mean weight less than 81grams is closest to:
Solution
To find the percentage of samples that have a sample mean weight less than 81 grams, we need to calculate the z-score for 81 grams and then find the area to the left of that z-score under the standard normal curve.
First, we calculate the standard error (SE), which is the standard deviation of the sampling distribution of the sample mean. The formula for the standard error is:
SE = σ / sqrt(n)
where σ is the population standard deviation and n is the sample size. Substituting the given values:
SE = 6.4 / sqrt(10) = 2.02
Next, we calculate the z-score for 81 grams. The formula for a z-score is:
Z = (X - μ) / SE
where X is the value from the dataset, μ is the population mean, and SE is the standard error.
Z = (81 - 85) / 2.02 = -1.98
Now, we need to find the area to the left of this z-score under the standard normal curve. This can be done using a standard normal distribution table or a statistical software or calculator. The value corresponding to Z = -1.98 in the standard normal distribution table is approximately 0.0244.
So, the percentage of samples that have a sample mean weight less than 81 grams is approximately 2.44%.
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