# Thread: z scores and probability

1. ## z scores and probability

I'm reading through the statistics chapter in my research methods textbook and think I may have found a mistake. I was hoping someone here could confirm that there's a problem. The mistake deals with using z scores to determine the probability results will fall between two values. Here are the basics of what the book says:

The two z scores are .72 and 1.74. The difference between the two scores is 1.02 or a little over one standard deviation. "We know that 68 percent of the values in an normal distribution lie between one standard deviation below the mean and one standard deviation above it, that is within a range of two standard deviations. One standard deviation therefore includes 34 percent of the values in a normal distribution, so we can project that approximately 34 percent of our sampled population" falls between the two values.

The two z scores were calculated using two responses from a survey about the number of hours spent on the internet every week. So the question is really dealing with the percentage of the sampled population that might fall between the two responses.

Based on what I've found, it's not as simple as just subtracting one z score from the other. You really need to look at a chart and then subtract the difference between the two values found there. In this case, I think the result should be around 19 percent rather than 34.

Am I missing something here? Is there any way the book could be correct? Thank you in advance for your responses.

2. Originally Posted by vw2000
I'm reading through the statistics chapter in my research methods textbook and think I may have found a mistake. I was hoping someone here could confirm that there's a problem. The mistake deals with using z scores to determine the probability results will fall between two values. Here are the basics of what the book says:

The two z scores are .72 and 1.74. The difference between the two scores is 1.02 or a little over one standard deviation. "We know that 68 percent of the values in an normal distribution lie between one standard deviation below the mean and one standard deviation above it, that is within a range of two standard deviations. One standard deviation therefore includes 34 percent of the values in a normal distribution, so we can project that approximately 34 percent of our sampled population" falls between the two values.

The two z scores were calculated using two responses from a survey about the number of hours spent on the internet every week. So the question is really dealing with the percentage of the sampled population that might fall between the two responses.

Based on what I've found, it's not as simple as just subtracting one z score from the other. You really need to look at a chart and then subtract the difference between the two values found there. In this case, I think the result should be around 19 percent rather than 34.

Am I missing something here? Is there any way the book could be correct? Thank you in advance for your responses.
The fraction of the normal distribution between 0.72 and 1.74 is 19%, as you have found. Quoting z-scores assumes the mean and the standard deviation are known already, representing the "sampled population." The procedures are different if you are asking different questions - for instance, you might want to know how likely it is that the two samples represent the same mean or not. But the projection that 34% lies between those z-scores is a serious blunder.

If the two results are being used to estimate a new mean and standard deviation, the statement would make more sense - but then the "new" z-scores would be -1 and +1, and the fraction between would be 68%.

3. ## Thank you

The mean and standard deviation were already known and shared earlier in the chapter. The author was trying to use one example about Internet usage and then apply several different statistical formulas to the same data.

In this specific section of the book, the author was only trying to show how z scores could be used to determine the probability that the sampled population would fall between two values.

He also referenced another example in which two z scores of 1.73 and -1.18 were used. In this example, he determined that since the range between the two is 2.91, that means that approximately 99 percent of the population's values lie between these two z scores (because it's close to three standard deviations). However, this also seems to be wrong. I'm getting 84 percent.

Considering these two examples, it seems he's just doing it wrong.

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