Statistics please help!!!?

As the owner of Phat-Wheel Repairs you rely on revenue from bike repairs to pay your bills. While the number of customers per week can vary, you choose to assume 25 bike repairs per-week for your calculations. Use X to denote the revenue per bike repair.
Suppose that X varies, having an expected value of $55 and a variance of ($)400 and a slightly right skewed distribution.

What is the probability that one weeks鈥?worth of your repair revenue will be
sufficient to pay your rent ($1200 per month)?

So you have to find the probability that your average repair will be worth $1200/25 = $48. We're saying that the expected average is $55 with a standard deviation of sqrt(400)=20.

So we want the probability that X>$48. But we need to standardize that score. The standard error is sigma/sqrt(n) = 20/5 = 4 (n=25). It says the distribution is slightly right-skewed, as in not normal. In addition, with n<30 we can't treat the mean as normally distributed, so we'll use a T-distribution (still not guaranteed reliable but should work well enough here).

T-score is ($48-$55)/4 = -1.75. So using a T-distribution table, the probability that T>-1.75 with 24 degrees of freedom is between .025 and .05. So the probability is 1 - that answer, so we get somewhere between .95 and .975.