Per Capita Income of Delaware Residents In a recent year, Delaware had the highest per capita annual income with $\$ 51,803$. If $\sigma=\$ 4850$, answer the following questions. Assume that the sample is taken from a large population and the correction factor can be ignored. Use a TI-83 Plus/TI-84 Plus calculator and round the answer to at least four decimal places.
Part 1 of 2
What is the probability that a random sample of 34 state residents had a mean income greater than $\$ 50,250$.
\[
P(\bar{X}> 50,250)=.9693
\]

Part: $1 / 2$

Part 2 of 2

What is the probability that a random sample of 34 state residents had a mean income less than $\$ 49,450$.
\[
P(\bar{X}< 49,450)=
\]

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 \(\boxed{0.0023}\).