First, let's use what we know, which is that the proportion of all first-year undergraduates in the US who are female (also known as the population proportion) is 56%, or 0.56 as a decimal.
Next, let's establish the size of our sample, which is going to be 200 students.
From this, we can calculate the mean and standard deviation for our normal approximation. The mean in this case is simply the population proportion, 0.56. To find the standard deviation, we use the formula for the standard deviation of a sample proportion, which is sqrt(p(1-p)/n), where p is the population proportion and n is the sample size. Filling in these numbers, we get sqrt(0.56(1-0.56)/200), which gives us approximately 0.0351.
Now, we can use these figures to calculate the z-scores for the proportions 0.50. A Z score is a measure of how many standard deviations an element is from the mean. To calculate the z-score for 50%, we subtract the mean from 0.50 and divide by the standard deviation, ((0.50 - 0.56) / 0.0351)), which gives us approximately -1.71. For less than 50%, it's equivalent to (0 - 0.56 / 0.0351), or about -15.95.
Next step is to calculate the probabilities of having more than 50% or less than 50% female students. For this, we need to apply the Cumulative Distribution Function to the respective z-scores. The probability that a simple random sample of 200 USFYU is over 50% female is calculated as 1 - CDF(-1.71), which gives a probability of about 0.956, or 95.6%. On the other hand, the probability of finding fewer than 50% are female is calculated as CDF(-15.95), which results in a virtually zero probability of approximately 1.33 x 10^-57.
In summary, based on our calculations, the probability that a simple random sample of 200 USFYU has more than 50% females is about 95.6%, and the probability that the sample has less than 50% females is virtually zero.