To find the probability that the mean of a sample of 100 observations drawn from a non-normal population will be less than 38, we can make use of the Central Limit Theorem (CLT). According to the CLT, the sampling distribution of the sample mean will be approximately normally distributed if the sample size is sufficiently large (and in this case, \(n = 100\) is considered large enough).
Given:
- Population mean (\(\mu\)) = 40
- Population standard deviation (\(\sigma\)) = 25
- Sample size (\(n\)) = 100
- Sample mean we are checking (\(\bar{x}\)) = 38
Step 1: Calculate the standard error (SE)
The standard error of the mean (SE) can be calculated using the formula:
\[ SE = \frac{\sigma}{\sqrt{n}} \]
Substituting the values we have:
\[ SE = \frac{25}{\sqrt{100}} = \frac{25}{10} = 2.5 \]
Step 2: Calculate the z-score
Next, we calculate the z-score for the sample mean of 38 using the formula:
\[ z = \frac{\bar{x} - \mu}{SE} \]
Substituting the values:
\[ z = \frac{38 - 40}{2.5} = \frac{-2}{2.5} = -0.8 \]
Step 3: Find the probability
Now we need to find the probability that the sample mean is less than 38, which corresponds to finding the cumulative probability for \(z = -0.8\).
Using standard normal distribution tables (or a calculator), we can find:
\[ P(Z < -0.8) \approx 0.2119 \]
Conclusion
Therefore, the probability that the mean of the sample will be less than 38 is approximately:
\[ \text{Probability} \approx 0.2119 \]
This means there is about a 21.19% chance that a random sample of 100 observations will have a mean less than 38.
1 answer