Question

the data value x exists in two data sets a and b. the mean is equal for both data sets. if the standard deviation for set a is greater than the standard deviation for set b, which is true for z_x for set a?

• it is equal to z_x for set b
• it is less than z_x for set b
• it is greater than z_x for set b

done

Explanation:

Step1: Recall z - score formula

The formula for the z - score of a data value \(x\) in a data set is \(z=\frac{x - \mu}{\sigma}\), where \(\mu\) is the mean of the data set and \(\sigma\) is the standard deviation of the data set.

Step2: Analyze the given conditions

We know that for data set \(A\) and data set \(B\), the mean \(\mu_A=\mu_B=\mu\) (since the mean is equal for both data sets) and the standard deviation \(\sigma_A>\sigma_B\). The data value \(x\) is the same for both data sets, so \(x - \mu\) is the same for both data sets (let's call this value \(d=x - \mu\)).

For data set \(A\), the z - score \(z_A=\frac{x-\mu}{\sigma_A}=\frac{d}{\sigma_A}\)

For data set \(B\), the z - score \(z_B=\frac{x - \mu}{\sigma_B}=\frac{d}{\sigma_B}\)

Since \(d\) is a constant (because \(x\) and \(\mu\) are the same for both sets) and \(\sigma_A>\sigma_B>0\) (standard deviation is non - negative, and since it's a measure of spread, for non - constant data sets \(\sigma>0\); if the data set has only one value, standard deviation is 0, but here we have a situation where \(\sigma_A\) and \(\sigma_B\) are being compared, so we can assume non - zero standard deviations), when we divide a non - zero constant \(d\) (if \(x eq\mu\); if \(x = \mu\), then \(z_A = z_B=0\)) by a larger positive number, the result is smaller.

Case 1: If \(x>\mu\), then \(d=x - \mu>0\). We have \(z_A=\frac{d}{\sigma_A}\) and \(z_B=\frac{d}{\sigma_B}\). Since \(\sigma_A>\sigma_B\), \(\frac{1}{\sigma_A}<\frac{1}{\sigma_B}\), and multiplying both sides by \(d>0\) gives \(\frac{d}{\sigma_A}<\frac{d}{\sigma_B}\), so \(z_A < z_B\).

Case 2: If \(x<\mu\), then \(d=x - \mu<0\). We have \(z_A=\frac{d}{\sigma_A}\) and \(z_B=\frac{d}{\sigma_B}\). Since \(\sigma_A>\sigma_B\), \(\frac{1}{\sigma_A}<\frac{1}{\sigma_B}\), and multiplying both sides by \(d<0\) (which reverses the inequality) gives \(\frac{d}{\sigma_A}>\frac{d}{\sigma_B}\)? Wait, no. Wait, if \(d\) is negative, let's take an example. Let \(d=- 2\), \(\sigma_A = 4\), \(\sigma_B = 2\). Then \(z_A=\frac{-2}{4}=-0.5\) and \(z_B=\frac{-2}{2}=-1\). Here, \(-0.5>-1\)? Wait, that's a contradiction. Wait, no, I made a mistake in the sign. Wait, if \(x<\mu\), then \(d=x - \mu\) is negative. So \(z=\frac{x - \mu}{\sigma}=\frac{\text{negative}}{\text{positive}}\) is negative. The magnitude of \(z_A\) is \(\frac{|d|}{\sigma_A}\) and the magnitude of \(z_B\) is \(\frac{|d|}{\sigma_B}\). Since \(\sigma_A>\sigma_B\), \(\frac{|d|}{\sigma_A}<\frac{|d|}{\sigma_B}\). So if \(d\) is negative, \(z_A=\ - \frac{|d|}{\sigma_A}\) and \(z_B=-\frac{|d|}{\sigma_B}\). Since \(\frac{|d|}{\sigma_A}<\frac{|d|}{\sigma_B}\), then \(-\frac{|d|}{\sigma_A}>-\frac{|d|}{\sigma_B}\)? Wait, no. Let's take numbers. Let \(d=- 4\), \(\sigma_A = 8\), \(\sigma_B = 2\). Then \(z_A=\frac{-4}{8}=-0.5\), \(z_B=\frac{-4}{2}=-2\). Here, \(-0.5>-2\), which means \(z_A>z_B\)? But this contradicts the previous case. Wait, what's wrong here?

Wait, no, the z - score formula is \(z=\frac{x-\mu}{\sigma}\). If \(x<\mu\), then \(x - \mu\) is negative, and \(\sigma\) is positive. So a more negative z - score means the data point is further to the left of the mean. Wait, in the first example, when \(x<\mu\), \(d=-2\), \(\sigma_A = 4\), \(z_A=-0.5\); \(\sigma_B = 2\), \(z_B=-1\). So \(z_A=-0.5\) is greater than \(z_B = - 1\) (because \(-0.5\) is to the right of \(-1\) on the number line). In the second example, \(d = - 4\), \(\sigma_A=8\), \(z_A=-0.5\); \(\sigma_B = 2\), \(z_B=-2\). Again, \(z_A=-0.5>z_B = - 2\). Wait, but in the case when \(x>\mu\), \(d = 2\), \(\sigma_A=4\), \(z_A = 0.5\); \(\sigma_B = 2\), \(z_B = 1\). So \(z_A=0.5<z_B = 1\).

Wait, so the key is: The z - score measures how many standard deviations a data point is away from the mean. If the standard deviation is larger, for a given distance from the mean (\(x-\mu\)), the number of standard deviations (the z - score) will be smaller in magnitude if \(x>\mu\) (since we are dividing by a larger \(\sigma\)) and larger in magnitude (less negative) if \(x<\mu\)? No, wait, magnitude: \(|z_A|=\frac{|x - \mu|}{\sigma_A}\), \(|z_B|=\frac{|x - \mu|}{\sigma_B}\). Since \(\sigma_A>\sigma_B\), \(|z_A|<|z_B|\). So if \(x>\mu\), \(z_A\) and \(z_B\) are positive, and \(z_A<z_B\) (because \(|z_A|<|z_B|\) and both positive). If \(x<\mu\), \(z_A\) and \(z_B\) are negative, and \(z_A>z_B\) (because \(|z_A|<|z_B|\) and a less negative number is greater than a more negative number). But wait, the problem says "the data value \(x\) exists in two data sets \(A\) and \(B\)". It doesn't specify if \(x\) is greater than or less than the mean. But wait, if \(x = \mu\), then \(z_A=z_B = 0\). But in general, let's think about the formula.

The z - score is \(z=\frac{x-\mu}{\sigma}\). Since \(\mu_A=\mu_B=\mu\) and \(x\) is the same, the numerator \(x - \mu\) is the same. The denominator \(\sigma_A>\sigma_B\). So when we divide the same numerator by a larger denominator, the result (the z - score) will be closer to zero (in magnitude) than when we divide by a smaller denominator.

If the numerator \(x-\mu\) is positive: \(z_A=\frac{x - \mu}{\sigma_A}\), \(z_B=\frac{x - \mu}{\sigma_B}\), \(\sigma_A>\sigma_B\) implies \(z_A<z_B\) (because dividing a positive number by a larger number gives a smaller result).

If the numerator \(x - \mu\) is negative: \(z_A=\frac{x - \mu}{\sigma_A}\), \(z_B=\frac{x - \mu}{\sigma_B}\), \(\sigma_A>\sigma_B\) implies \(z_A>z_B\) (because dividing a negative number by a larger number gives a result that is less negative, i.e., greater, than dividing by a smaller number). But wait, this seems contradictory. But the problem is asking "which is true for \(z_x\) for set \(A\)". Wait, maybe there is a mistake in my second case. Wait, no, let's take the formula again. Let's assume that \(x eq\mu\) (if \(x = \mu\), then \(z_A=z_B = 0\)).

Wait, the question is about the relationship between \(z_A\) and \(z_B\). Let's use the formula \(z=\frac{x - \mu}{\sigma}\). Let \(k=x - \mu\) (constant for both sets). Then \(z_A=\frac{k}{\sigma_A}\), \(z_B=\frac{k}{\sigma_B}\). Since \(\sigma_A>\sigma_B>0\), we can write \(\frac{z_A}{z_B}=\frac{\frac{k}{\sigma_A}}{\frac{k}{\sigma_B}}=\frac{\sigma_B}{\sigma_A}\) (assuming \(k eq0\)). Since \(\sigma_A>\sigma_B\), \(\frac{\sigma_B}{\sigma_A}<1\). So \(\frac{z_A}{z_B}<1\). If \(k>0\), then \(z_A\) and \(z_B\) are positive, so \(z_A<z_B\) (because \(\frac{z_A}{z_B}<1\) and both positive). If \(k < 0\), then \(z_A\) and \(z_B\) are negative, and \(\frac{z_A}{z_B}<1\) implies \(z_A>z_B\) (because if \(a\) and \(b\) are negative and \(\frac{a}{b}<1\), then \(a>b\); for example, \(a=-1\), \(b = - 2\), \(\frac{-1}{-2}=0.5<1\) and \(-1>-2\)). But the problem doesn't specify the sign of \(k=x - \mu\). However, in the context of z - scores, the most common interpretation is that we are looking at the relative position. But wait, maybe the problem assumes that \(x\) is not equal to the mean (otherwise, both z - scores are zero). But let's check the options. The options are "It is equal to \(z_x\) for set \(B\)", "It is less than \(z_x\) for set \(B\)", "It is greater than \(z_x\) for set \(B\)".

Wait, maybe I made a mistake in the second case. Wait, no, let's think of the standard deviation as a measure of spread. A larger standard deviation means the data is more spread out. So a data point \(x\) that is a certain distance from the mean will be a "smaller" z - score (in magnitude) in a more spread - out data set. For example, suppose the mean is 10, \(x = 12\) (so \(x-\mu = 2\)). For set \(A\) with \(\sigma_A = 4\), \(z_A=\frac{2}{4}=0.5\). For set \(B\) with \(\sigma_B = 2\), \(z_B=\frac{2}{2}=1\). So \(z_A = 0.5<z_B = 1\). If \(x = 8\) (so \(x-\mu=- 2\)), for set \(A\) with \(\sigma_A = 4\), \(z_A=\frac{-2}{4}=-0.5\). For set \(B\) with \(\sigma_B = 2\), \(z_B=\frac{-2}{2}=-1\). Here, \(z_A=-0.5>z_B=-1\). But this is a problem. But the options don't have a "it depends" option. So maybe the problem assumes that \(x>\mu\) or that we are considering the magnitude? Wait, no, the z - score can be positive or negative. But the options are "less than", "greater than", or "equal".

Wait, maybe the key is that the formula is \(z=\frac{x - \mu}{\sigma}\), and since \(\sigma_A>\sigma_B\), and the numerator is the same, when we divide by a larger number, the result is smaller in absolute value, but the sign depends on \(x-\mu\). But maybe the problem is considering the case where \(x-\mu\) is positive (the most intuitive case, as z - scores for values above the mean are positive). In that case, \(z_A<z_B\).

Alternatively, maybe there is a mistake in my second case analysis. Wait, no, let's go back to the formula. The z - score is a measure of how many standard deviations a data point is from the mean. If the standard deviation is larger, for the same distance from the mean, the number of standard deviations (the z - score) is smaller. For example, if you are 2 units above the mean, in a data set with standard deviation 4, you are 0.5 standard deviations above the mean (z = 0.5). In a data set with standard deviation 2, you are 1 standard deviation above the mean (z = 1). So here, \(z_A = 0.5<z_B = 1\). If you are 2 units below the mean, in a data set with standard deviation 4, you are - 0.5 standard deviations (z=-0.5), and in a data set with standard deviation 2, you are - 1 standard deviations (z = - 1). Here, - 0.5 is greater than - 1 (because - 0.5 is to the right of - 1 on the number line). But the problem doesn't specify if \(x\) is above or below the mean. However, the options include "It is less than \(z_x\) for set \(B\)", which is true when \(x>\mu\), and "It is greater than \(z_x\) for set \(B\)" which is true when \(x<\mu\). But since the problem is likely considering the general case or the case where \(x eq\mu\) and the most common interpretation (positive z - score when \(x>\mu\)), the answer is that \(z_x\) for set \(A\) is less than \(z_x\) for set \(B\).

Answer:

It is less than \(z_x\) for set \(B\)