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z =0 and P(X< x) = 0.5 Explanation: z = (x-xbar)/sd, where xbar is the estimated mean or average of the sample, sd is the standard deviation, and x is the value of the particular outcome. We change x to z so that we can use the normal distribution or t-tables tables, which are based on a zero mean and 1 standard deviation. For example: What is the probability that the mean value of the distribution is 5 or less, given the sample average is 5 and the sd is 2? The z-score would be (5-5)/2 which is equal to 0. The probability, if we assume the normal or t-distribution, is 0.50. (see normal distribution tables) I hope this makes sense to you. The normal distribution is symmetrical. Per the example, a sample average of 5 tells you there is equal chance of the population mean being above and below 5.

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