Final answer:
In a standard normal distribution, the probability that z is greater than zero is 0.5 or 50$, due to the symmetry of the distribution around the mean of 0.
Step-by-step explanation:
In a standard normal distribution, the mean is 0 and the standard deviation is 1. This distribution, denoted as Z~ N(0, 1), is symmetric about the mean. Therefore, the probability that z is greater than zero is equal to the probability that z is less than zero since the total area under the curve is 1 (representing 100$ probability).
As the standard normal distribution is symmetric, half of the area under the curve lies to the right of the mean, and half to the left. Hence, the probability that z is greater than zero is 0.5 or 50$.
To find probabilities associated with specific z-scores, one can use a z-table, a calculator, or computer software that provides areas under the normal curve. For example, if we look up a z-score of 1 in a z-table, we would find that the area to the left of z 1 (or the cumulative probability up to 1) is approximately 0.8413. Therefore, the area to the right, which represents the probability that z is greater than 1, would be 1 - 0.8413, which is approximately 0.1587, or 15.87$.