To find the probability that the mean of the sample will exceed 45, we can use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of the sample mean will be approximately normally distributed if the sample size is sufficiently large (which it is, given n = 100).
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Determine the parameters of the sampling distribution:
- Population mean (\(\mu\)) = 40
- Population standard deviation (\(\sigma\)) = 25
- Sample size (\(n\)) = 100
The mean of the sampling distribution of the sample mean (\(\mu_{\bar{x}}\)) is the same as the population mean: \[ \mu_{\bar{x}} = \mu = 40 \]
The standard deviation of the sampling distribution of the sample mean (also known as the standard error, \(SE\)) is given by: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{25}{\sqrt{100}} = \frac{25}{10} = 2.5 \]
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Standardize the sample mean using the Z-score formula: We want to find the probability that the sample mean (\(\bar{x}\)) exceeds 45. We first calculate the Z-score: \[ Z = \frac{\bar{x} - \mu_{\bar{x}}}{SE} = \frac{45 - 40}{2.5} = \frac{5}{2.5} = 2 \]
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Find the probability corresponding to the Z-score: We need to find the probability that \(Z > 2\). Using the standard normal distribution table or a calculator:
- The probability that \(Z < 2\) is approximately 0.9772 (this is the cumulative probability up to Z = 2).
- Therefore, the probability that \(Z > 2\) is: \[ P(Z > 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228 \]
Thus, the probability that the mean of the sample will exceed 45 is approximately 0.0228, or 2.28%.