Find the probability that the mean of a random sample of 25 elements from a normally
distributed population with a mean 90 and standard deviation of 60 is larger than 100.
To find the probability that the mean of a random sample is larger than a certain value, we can use the Central Limit Theorem and the z-score.
The Central Limit Theorem states that if we take a random sample from any population with a large enough sample size, the sample mean will be approximately normally distributed.
To solve this problem, we will standardize the distribution using the z-score.
The formula for the z-score is:
z = (x - μ) / (σ / √n)
Where:
- x is the given value (in this case, 100)
- μ is the population mean (90)
- σ is the population standard deviation (60)
- n is the sample size (25)
Substituting the values in the formula,
z = (100 - 90) / (60 / √25)
z = 10 / (60 / 5)
z = 10 / 12
z ≈ 0.8333
Now, we need to find the probability that the z-score is greater than 0.8333. We can do this by looking up the z-table or using a calculator that has a built-in function to find the area under the standard normal curve.
Using the z-table, we find that the probability corresponding to a z-score of 0.8333 is approximately 0.7967.
Therefore, the probability that the mean of a random sample of 25 elements from a normally distributed population with a mean of 90 and a standard deviation of 60 is larger than 100 is approximately 0.7967 or 79.67%.