Step 1 :We are given that the population mean (μ) is $51,803, the standard deviation (σ) is $4,850, and the sample size (n) is 34. We are asked to find the probability that a random sample of 34 state residents had a mean income less than $49,450.
Step 2 :We can use the formula for the standard error of the mean to calculate the z-score. The formula is \((X - μ) / (σ / √n)\), where X is the sample mean.
Step 3 :Substituting the given values into the formula, we get \(z = (49450 - 51803) / (4850 / √34)\), which simplifies to \(z = -2.828913362591957\).
Step 4 :We can then use the z-score to find the probability. The probability corresponding to this z-score is 0.002335317049737719.
Step 5 :Rounding this to four decimal places, we get 0.0023.
Step 6 :Thus, the probability that a random sample of 34 state residents had a mean income less than $49,450 is \(\boxed{0.0023}\).