Let’s talk standard deviation. Your first thought is to start assembling a list of increasingly frantic questions:
Do we need to know that horrible formula I learned in Stats class? (No.)
Do we need to know the relationship between variance and Standard deviation? (You just need to know that there is a relationship, and that if you can solve for one, you can solve for the other.) Etc.
So, rather than droning on about what we don’t need to know, let’s boil down what we do need to know about standard deviation. The good news – it isn’t much. Just make sure you’ve internalized the following:
- The standard deviation is a measure of the dispersion the elements of the set around mean. The farther away the terms are from the mean, the larger the standard deviation.
- If we were to increase or decrease each element of the set by “x,” the standard deviation would remain unchanged.
- If we were to multiply each element of the set by “x,” the standard deviation would also be multiplied by “x.”
- If the mean of a set is “m” and the standard deviation is “d,” then to say that something is within 3 standard deviations of a set is to say that it falls within the interval of (m – 3d) to (m + 3d.) And to say that something is within 2 standard deviations of the mean is to say that it falls within the interval of (m – 2d) to (m + 2d.)
That’s basically it. Not anything to get too worked up about. So, let’s see some of these principles in action to substantiate the claim that we won’t have to do too much arithmetical grinding on these types of questions:
If d is the standard deviation of x, y, z, what is the standard deviation of x+5, y+5, z+5 ?
A) d
B) 3d
C) 15d
D) d+5
If our initial set is x, y, z, and our new set is x+5, y+5, and z+5, then we’re adding the same value to each element of the set. We already know that adding the same value to each element of the set does not change the standard deviation. Therefore, if the initial standard deviation was d, the new standard deviation is also d. We’re done – the answer is A. (You can see this with a simple example. If your initial set is {1, 2, 3} and your new set is {6, 7, 8} the dispersion of the set clearly hasn’t changed.)
Surely the questions get harder than this, you say. They do, but if you know the aforementioned core concepts, they’re all quite manageable. Here’s another one:
Some water was removed from each of 6 tanks. If standard deviation of the volumes of water at the beginning was 10 gallons, what was the standard deviation of the volumes at the end?
1) For each tank, 30% of water at the beginning was removed
2) The average volume of water in the tanks at the end was 63 gallons
We know the initial standard deviation. We want to know if it’s possible to determine the new standard deviation after water is removed. To the statements we go!
Statement 1: If 30% of the water is removed from each tank, we know that each term in the set is multiplied by the same value: 0.7. Well, if each term in a set is multiplied by 0.7, then the standard deviation of the set is also multiplied by 0.7. If the initial standard deviation was 10 gallons, then the new standard deviation would be 10*(0.7) = 7 gallons. And we don’t even need to do the math – it’s enough to see that it’s possible to calculate this number. Therefore, Statement 1 alone is sufficient.
Statement 2: Knowing the average of a set is not going to tell us very much about the dispersion of the set. To see why, imagine a simple case in which we have two tanks, and the average volume of water in the tanks is 63 gallons. It’s possible that each tank has exactly 63 gallons and, if so, the standard deviation would be 0, as everything would equal the mean. It’s also possible to have one tank that had 126 gallons and another tank that was empty, creating a standard deviation that would, of course, be significantly greater than 0. So, simply knowing the average cannot possibly give us our standard deviation. Statement 2 alone is not sufficient to answer the question.
And the answer is A.
Maybe at this point you’re itching for more of a challenge. Let’s look at a slightly tougher one:
7.51; 8.22; 7.86; 8.36
8.09; 7.83; 8.30; 8.01
7.73; 8.25; 7.96; 8.53
A vending machine is designed to dispense 8 ounces of coffee into a cup. After a test that recorded the number of ounces of coffee in each of 1000 cups dispensed by the vending machine, the 12 listed amounts, in ounces, were selected from the data above. If the 1000 recorded amounts have a mean of 8.1 ounces and a standard deviation of 0.3 ounces, how many of the 12 listed amounts are within 1.5 standard deviations of the mean?
A)Four
B) Six
C) Nine
D) Eleven
Okay, so the standard deviation is 0.3 ounces. We want the values that are within 1.5 standard deviations of the mean. 1.5 standard deviations would be (1.5)(0.3) = 0.45 ounces, so we want all of the values that are within 0.45 ounces of the mean. If the mean is 8.1 ounces, this means that we want everything that falls between a lower bound of (8.1 – 0.45) and an upper bound of (8.1 + 4.5). Put another way, we want the number of values that fall between 8.1 – 0.45 = 7.65 and 8.1 + 0.45 = 8.55.
Looking at our 12 values, we can see that only one value, 7.51, falls outside of this range. If we have 12 total values and only 1 falls outside the range, then the other 11 are clearly within the range, so the answer is D.
As you can see, there’s very little math involved, even on the more difficult questions.
Takeaway: remember the axiom that the more complex-looking the formula is for a concept, the simpler the calculations are likely to be on the CAT. An intuitive understanding of a topic will always go a lot further on this test than any amount of arithmetical virtuosity.
