Answer: Calculating a z score and p

Suppose that SAT scores among U.S. college students are normally distributed with a mean of 500 and a standard deviation of 100. What is the probability that a randomly selected individual from this population has an SAT score at or below 600?

Using the formula to calculate the z value, we find z = (x – μ)/σ = (600 – 500)/100 = +1.00. Recall that the z-score is the number of standard deviations that the score of interest differs from the mean. A score of 600 is one standard deviation above the mean, so it has a z value of 1.00.

Using either a z table or the pz converter, we find that the probability that a randomly selected z score in a normal distribution will exceed z = 1.00 is .159, the right-tailed p value, or about 16%.

The probability that z will exceed 1.00 for a randomly selected score is equivalent to the probability that a randomly selected individual from this population will have an SAT score over 600, about 16%. The probability that a randomly selected individual from this population will have an SAT score at or below 600 is 100% – 16% = 84%, which is the left-tailed p value, .841, as shown in the graphical version of the pz calculator (click on the “Graphic” button and enter in a mean of 500, a standard deviation of 100, and a raw score of 600–be sure to press “Enter” or “Return” on the keyboard after each entry).

Correction for continuity: When we have a distribution that is not truly continuous, we can improve the estimate of probability. In the case of SAT scores, possible values are in increments of 10. Possible scores are 590, 600, 610, etc., but not 603.214…. etc. Thus, a score of 600 represents any score between 595.000… and 605.000…. on an underlying continuous distribution. To compute the probability of a score at or below 600, we should use the value that divides 600 from the next possible score, 610, giving us 605 as the real limit. Now, the corrected z value is z = (x – μ)/σ = (605 – 500)/100 = 105/100 =  +1.05.  The probability that a randomly selected z score in a normal distribution will exceed z = 1.05 is .147, the right-tailed p value, or about 15%.

The correction could be important in some situations. It is appropriate to be applied any time we are using a measure that is not truly continuous, though it may not make much difference.  In this example, it would be wrong to report the original answer of .159 when the correct answer is .147.

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