Question Video: Identifying a Set of Values with Zero Dispersion | Nagwa Question Video: Identifying a Set of Values with Zero Dispersion | Nagwa

Question Video: Identifying a Set of Values with Zero Dispersion Mathematics • Third Year of Preparatory School

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If the dispersion of a set of values is equal to zero, then which of the following are true? [A] The difference between the individuals is great. [B] The difference between the individuals is small. [C] All the values are equal. [D] The arithmetic mean of these values is zero. [E] All the values are negative.

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Video Transcript

If the dispersion of a set of values is equal to zero, then which of the following are true? (a) The difference between the individuals is great. (b) The difference between the individuals is small. (c) All the values are equal. (d) The arithmetic mean of these values is zero. Or (e) all the values are negative.

The dispersion of a data set can be measured using the standard deviation, which we denote by 𝜎 sub π‘₯. For a data set containing the 𝑛 values π‘₯ one, π‘₯ two, π‘₯ three, and so on all, the way up to π‘₯ 𝑛, with a mean value equal to πœ‡, the standard deviation is calculated using the following formula. 𝜎 sub π‘₯ is equal to the square root of π‘₯ one minus πœ‡ squared plus π‘₯ two minus πœ‡ squared, and so on, all the way up to π‘₯ 𝑛 minus πœ‡ squared divided by 𝑛. Practically, we subtract the mean value from each π‘₯-value, square this, find the sum, divide by 𝑛, that’s the number of values there are in the data set, and finally take the square root.

If the dispersion of a data set is equal to zero, then the standard deviation is equal to zero. So we have that zero is equal to the formula we’ve previously wrote down. Squaring both sides of this equation and then multiplying by 𝑛, we have that zero is equal to π‘₯ one minus πœ‡ all squared plus π‘₯ two minus πœ‡ all squared, and so on, up to plus π‘₯ 𝑛 minus πœ‡ all squared. We now have that the sum of these 𝑛 expressions is equal to zero. But we know that if we square any real number, regardless of whether it is positive, negative, or equal to zero, we get a value greater than or equal to zero. And so in order for the sum of these nonnegative values to be equal to zero, it must be the case that each set of parentheses is itself exactly equal to zero.

So, setting each set of parentheses equal to zero, we have π‘₯ one minus πœ‡ equals zero, π‘₯ two minus πœ‡ equals zero, and so on all the way down to π‘₯ 𝑛 minus πœ‡ is equal to zero. Adding πœ‡ to each side of each equation, we then have that π‘₯ one is equal to πœ‡, π‘₯ two is equal to πœ‡, all the way down to π‘₯ 𝑛 is equal to πœ‡. And so all the values in the data set are equal to the mean and equal to one another. This is option (c), all the values are equal. The dispersion of the values is equal to zero because there is no spread about the mean. All of the values are individually equal to the mean.

Let’s just briefly consider each of the other options. Options (a) and (b) describe the difference between the individuals, either as great or as small. But as we’ve just seen, if all of the values are equal, then there is no difference between the individuals. Option (d) says that the arithmetic mean of these values is zero. Well, this doesn’t have to be the case. As long as all the values are equal, then they’ll all be equal to the arithmetic mean πœ‡, but it doesn’t matter what the value of this mean is. And finally, option (e) says that all of the values are negative. Well again, this doesn’t have to be the case. It could be the case that all of the values are negative, but they could all be positive, or indeed they could all be equal to zero. As long as they’re all equal to each other, the dispersion will be equal to zero.

We found then that if the dispersion of a set of values is equal to zero, then all of the values are equal.

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