The mean rent of an apartment is $780. You randomly select 9 apartments. Compute the probability that the mean rent is less than $825.?

Mean μ = 780
Standard deviation σ = 150
Standard error σ / √ n = 150 / √ 9 = 50
standardize xbar to z = (xbar - μ) / (σ / √ n )
P(xbar < 825) = P( z < (825-780) / 50)
= P(z < 0.9) = 0.8159 <---------- probability that the mean rent is less than $825
(from normal probability table)

Unable to answer this question without additional data. From what distribution? Is it normally distributed? What is the standard deviation? Do you know how to compute a z-score? How to use a z-table?

After update:
“Assume that the rents are normally distributed with a mean of $780 and standard deviation of $150.

You randomly select 9 apartments. Compute the probability that the mean rent is less than $825.”

Now you know the distribution of the means: normal with mean 780 and standard deviation 150/√9 = 150/3 = 50.

Z-score is (825-780)/50 = 45/50 = +.90

From a z table
P(Z < .90) = .81594 ~ .82 ~ 82%