A population has a mean of 200 and standard deviation of 50 and simple random sample size of 100 and x is used to estimate u. What is the probility that the sample mean will be within plus or minue 5 of the population mean?
To calculate the probability that the sample mean will be within plus or minus 5 of the population mean, we can use the Central Limit Theorem (CLT).
The CLT states that for a large enough sample size, the distribution of sample means will be approximately normally distributed, regardless of the shape of the population distribution.
In this case, the sample size is 100, which is large enough to apply the CLT.
The mean of the sample means is equal to the population mean (μ) and the standard deviation of the sample means (also known as the standard error) is equal to the population standard deviation (σ) divided by the square root of the sample size (n).
To calculate the standard error (SE), we use the formula:
SE = σ / √n
In this case, the population standard deviation (σ) is 50 and the sample size (n) is 100.
SE = 50 / √100 = 50 / 10 = 5
So, the standard error is 5.
Now, we can calculate the z-scores for the upper and lower bound of the sample mean:
Z_upper = (X_upper - μ) / SE
Z_lower = (X_lower - μ) / SE
Where X_upper is the upper bound of the sample mean, X_lower is the lower bound of the sample mean, and μ is the population mean.
In this case, X_upper = μ + 5 = 200 + 5 = 205
X_lower = μ - 5 = 200 - 5 = 195
So, Z_upper = (205 - 200) / 5 = 5 / 5 = 1
Z_lower = (195 - 200) / 5 = -5 / 5 = -1
Now, we can use a standard normal distribution table or a calculator to find the probability that a z-score falls between -1 and 1. This is equivalent to finding the area under the curve between -1 and 1.
Using the standard normal distribution table, we find that the probability of a z-score falling between -1 and 1 is approximately 0.6827.
Therefore, the probability that the sample mean will be within plus or minus 5 of the population mean is approximately 0.6827 or 68.27%.