Statistics Flashcards

1
Q

Average/Sum of Terms Formula:

A

Formula for Average of set of numbers:

Average = Sum/Number of Terms

If you have a long list of numbers and you need to find the sum of the terms, for instance in a sequence of numbers, use this variation of the average formula:

Sum of Terms = Average x Number of Terms

So you find out the Average and the number of terms and that’s a quick way of figuring out the sum of terms.

REMEMBER: If average is unknown use Average Formula. If Average is known use Sum Formula.

E.g. Sum of 6 numbers is 90. What is average?

A = Sum/No. of Terms = 90/6 = 15

REMEMBER: you don’t need all numbers in a set to find the average. Think critically to figure it out by using just the Average Formula.

In more complex average problems you have to set up two average formulas.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Complex Average Problems:

A

In complex average problems you may have to set up two average formulas. Set up table to keep track of two formulas:

                Average     x     Number (sales)   =   Sum

Old Total 800 n 800n

This Sale $2000 1 2000

New Total 900 n + 1 900(n+1)

In this example we’re looking for number of sales. Note how the Number and Sum columns add up but values in Average column don’t.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

Median:

A
  1. If an odd group of numbers is arranged in numerical order, the median is the middle value after the terms are arranged in numerical order.

E.g. What is median of 4, 5, 100, 1, 6?

 arranged in numerical order: 1, 4, 5, 6, 100

   The median: 5
  1. If an even group of numbers is arranged in numerical order, the median is the average of the two middle terms after the terms are arranged in numerical order:

E.g. What is median in 2, 9, 8, 17, 11, 37?

arranged in numerical order: 2, 8, 9, 11, 17, 37

two middle terms are 9 and 11 and the average is:

                     9 + 11/2 = 10

Caution: Don’t confuse median with average or arithmetic mean which are different things. To find the average you calculate the sum of all terms and then multiply by the number of terms. In consecutive sets though median and average are equal.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Mode:

A

The mode is the number that appears most frequently in a set of numbers.

E.g. in the set 1, 2, 2, 2, 3, 4, 4 the mode is 2.

If the frequency is the same for two numbers both numbers are the mode because a set can have two modes.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

Range:

A

The range is the positive difference between the smallest term and the largest term in a set of numbers:

E.g. In the set 2, 4, 10, 20, 26, the range is 24.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

Standard Deviation:

A

The standard deviation represents how close or far the terms in a list of numbers are from the average of the numbers.

E.g. the sets 1, 2, 3 and 101, 102, 103 have the same standard deviation since they both have one term on the average and two terms exactly one unit away from the average.

But for instance set 1, 3, 5 has a smaller standard deviation than 0, 3, 6 because while both have the same average, 3, in the first set there are terms 2, 0, and 2 units from the average, in the second one there are terms 3, 0, and 3 units from the terms. So the standard deviation is larger in the set with more widely spaced numbers.

A small SD indicates that set is clustered closely around the average.

A large SD indicates that the set is spread out widely, with some points far from the average.

Finding SD:

Use critical thinking and quick sketches of number lines to determine how close the terms in a list are from the average. If you understand how to figure out the average spread of a set, you will be able to answer all GMAT SD questions.

E.g. set {2, 4, 6, 8}

Difference from mean: {3, 1, 1, 3}

Average spread: 2

on GMAT you will most likely have to figure out one of these two things:

  1. Changes in SD when a set is transformed.
  2. Comparisons of SDs of multiple sets.

Remember for that:

  • if problem on changes in SD, ask yourself if changes are moving the data closer to mean, farther, or neither.
  • if problem on comparison, ask yourself which set is more spread out from its mean.

REMEMBER These:

  • An SD of 0 means: all numbers in set are equal.
  • If each data point in a set is increased by a certain number (i.e. a certain number added to each data point) the SD remains unchanged.
  • If each data point in a set in changed by a factor (i.e. multiplied by a number) the SD will increase because it will make all the gaps bigger by the factor of that number and so each number will be farther away from mean by that factor (e.g. if increased by factor of 7, gaps will become 7 times bigger; SD will increase by factor of 7).
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

Average and Median of Consecutive Integers:

A

In a set of consecutive, or evenly spaced, integers the average equals the median.

E.g. in 1, 2, 3, 4, 5,
both the average and the median are 3.

A faster way of determining the average in a set of consecutive integers: It’s the average of the first and last terms in the set.

I.e. 1 + 5/2 = 6/2 = 3

When there are an odd number of members in a consecutive set, the mean and median will be a member of the set and thus an integer.

E.g. 5,6,7,8,9 mean and median = 7

When there are an even number of members in a consecutive set, the mean and median will not be a member of the set and not an integer.

E.g. 5,6,7,8 mean and median = 6.5

REMEMBER: Whenever you see a question on consecutive integers on GMAT check if you need to use the Average formula or the formula for the sum of consecutive integers. Most likely you will have to use one of the two.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Weighted Average:

A

Average Formula applies if all individual values are equally weighted, i.e. they all count equally towards the average. But in weighted averages, the values are weighted differently, some more heavily than others. Weighted averages therefore include the individual weightings. So, you’ll always have a component multiplied by its weighting, and then added to that is the next component multiplied by its weighting etc. Depending on where the weighting is heavier you can often see, what value the weighted (overall) average is going to be closer to.

Weighted Average = (Percent1)x(Average1) + (Percent2)x(Average2)…

This formula works for all problems in which two entities, each with a different proportion or average, are combined to form a total with an overall proportion or average that’s different from its individual components.

GMAT often just sticks to 2 components in these questions.

E.g. You scored 100 points in final exam which weight 3/5 into final grade, and 80 points in another exam which weighs 2/5. What is the average?

100x3/5 + 80x2/5 = 60 + 32 = 92

REMEMBER: The normal average between two points would be calculated using the normal average formula (a = sum/# terms). That average will always be exactly between the two points. And depending on which side has more weight (more than 50%) the final average will be closer to that number and less or more than normal average.

Another way of solving. You know which score is weighted more. You also know that there are 20 points in between 100 and 80. So take 3/5 of 20. That’s 12. Then go from the lighter side (80) 12 point up and reach 92. If the other test was weighted heavier, you would go from the other side and count down (100-12 = 88)

Try out and decide which of the two methods you like better.

If you were only given the scores and asked to find the proportions or weights:

E.g. You scored 100 points in final exam, and 80 points in another exam. Overage score ends up being 92. How was your final score weighted?

What you do to solve:

  1. You still know there are 20 between 100 and 80.
  2. You also know that the 100 points (final exam) were weighted heavier, so go from the other end (80) and calculate how many point are between 80 and the final score of 92. Then divide that by the entire difference of 20 and you got your proportion.

12/20 = 3/5 which is 60%.

If you were asked for the proportion of the other exam you would go form the other end (100) because you know the 80-point exam was weighted lighter.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Mixtures on GMAT:

A

Mixtures on GMAT are most commonly mixtures of liquids. They seem intimidating at first, but really they are just proportion, ratio, and percentage problems in disguise. Most mixture questions will ask you to combine two portions that are themselves subdivided into portions.

For mixture problems, translate the given information into ratios.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

Balanced Average Approach on Mixture Problems:

A

On mixture problems where you are asked how much of one mixture and of another mixture you have to mix to make a specific mix you can use the balance average approach.

E.g. How much of substance x that has 20% bleach and substance y that has 45% bleach do you have to mix for a final mixture that has 35% bleach?

There are two ways to solve this:

Algebra Way:

The normal average between 20% and 45% would be 32.25%. So, this shows you that 35% is closer to 45% so it has more of the substance y.

Set up equation and solve for x/y:

  1. 2x + 0.45y = 0.35(x + y)
  2. 2x + 0.45y = 0.35x + 0.35xy
  3. 1y = 0.15x
  4. 1/0.15 = x/y

2/3 = x/y

  1. Now you know that in order to make a mixture with 35% bleach you need 2 parts x and 3 parts y. But you are asked how much of the final mixture will be made of substance x. So now you have to find the number for the whole:
    2: 3:5
  2. Now you know that the ratio of substance x to the final mixture is 2:5 or 2/5. that means 2/5 of the final mixture will be made of substance x. And for GMAT test day you should know that 2/5 = 40%.

Teeder-Totter Way:

Draw teeter totter and you will see that the middle point between 20 and 45 and therefore normal average is exactly 32.5. But you want to get to 35%. You know the teeter totter will be tilted toward y as 35 is closer to 45 than to 20. So as always with teeter totter begin counting from the lighter side, side x. In order to go from side x (20) to 35 you are counting 15 points (that is y). Then in order from the y side (45) to get to 35 you are counting 10 points (that is x). So your ratio of x/y will be 10/15 = 2/3. REMEMBER: Don’t mix y and x up when you do the teeter totter method. If you begin on the lighter side you are counting the weight of the opposite side, the heavier-weight side, in this case you start counting from side x and the points you count from there to get to 35 (that is 15) is the weight for y.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

Volume Equation for Mixture Problems:

A

If you have a mixture problem where one substance stays the same and one is added to dilute the concentration, you can make the calculation easy by setting up an equation where you multiply the volume (e.g. certain liters of water) by the percent concentration (e.g. of sugar in the water)

Volume of Sugar = Volume Water x Sugar Concentration

so if 10% of 10 Liters is sugar then:

Volume of Sugar = 10/100 x 10

Because the volume of sugar doesn’t change if you dilute the mixture by adding water you set the old and new expressions equal to each other to find out the new Volume:

Volume1 x Concentration = Volume2 x Concentration.

Volume two is what you’re looking for in these mixture questions. So, if for instance the concentration of sugar in the new mixture should be 4% you can write:

10/100 x 10 = 4/100 x V2

And then solve for V2.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Consecutive Integers

A

A sequence is an ordered list of numbers a(n-1), an, a(n+1) where n denotes where in the sequence the number is located. Note that the expression n-1, n, n+1 etc. goes slightly under the base a.

E.g. if n = 2 a2 is the 2nd number
a(2+1) is a3 which is the third number
a(2-1) is a1 which is the first number

Picking numbers is a useful strategy in sequence questions if you’re not given concrete values for the terms in the sequence.

Remember: If the question stem talks about “consecutive sequence” you should know that in a consecutive sequence half the numbers must be greater and half the numbers must be less than the mean. So if you are given a number and know it’s a consecutive sequence, first figure out the average (if a sequence of 4 consecutive numbers has the sum of 268 you know the average is 268/4 = 67. Then you know that 67 is in the middle of that sequence of 4, with two ahead and two after. Let’s say the question stem said the numbers are even integers. That means you know the four numbers are:

64, 66, 68, 70

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

Different Types of Consecutive Integers:

A
  1. Consecutive Integers: All values in set increase by 1. (E.g. {12, 13, 14, 15, 16})
  2. Consecutive Multiples: All values in set are multiples of an increment. (E.g. {12, 16, 20, 24} each value increases from one to next by 4 and each has multiple 4 which is the increment)
  3. Consecutive Even Integers: E.g. {8, 10, 12, 14}. Incorrect: {8, 10, 14, 16} because skips 12.
  4. Consecutive Primes: E.g. {11, 13, 17, 19}
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Number of Terms in Sequence of Consecutive Integers:

A
  1. In sets of consecutive integers take difference between largest and smallest and add 1.

Formula:
Last – First + 1

E.g. if 89 is largest 36 is smallest:

89 - 36 + 1 = 53

  1. In sets of consecutive multiples you can’t use the formula above because you would be overcounting because you are not considering the increment. Here you have to divide by increment before adding 1.

Formula:
((Last - First)/Increment) + 1

Increment could for instance be 4 if all values in set increase by 4 and are multiples of 4. The bigger the increment, the smaller the result. Use this formula for questions like this:

How many multiples of 7 are there between 100 and 150?

First search for first and last multiple of 7 in the sequence.

147-105/7 + 1 = 42/7 + 1 = 6 + 1 = 7

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

Properties of Evenly Spaced Sets:

A
  1. Arithmetic mean and median are equal in evenly spaces sets.

E.g. {4, 8, 12, 16, 20, 24} mean and median = 4+24/2 = 14

  1. Mean and median of evenly spaced sets are equal to the average of First and Last terms. For all evenly spaced sets Average = Last - First/2. (so you can calculate the average in evenly spaced sets by adding the middle valued, or the first and last and then dividing the respective number by 2)

E.g. {4, 8, 12, 16, 20}
mean and median = 4+20/2 = 12

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Sum of Consecutive Integers:

A

E.g. Sum of all integers from 20 to 100 inclusive?

Steps:

  1. Formula for Sum in evenly spaced sets:

Sum = Average x Number of Terms

  1. Calculate Average:

20 + 100/2 = 60

  1. Calculate Number of Terms with Formula:

Last-First + 1 = 100 - 20 + 1 = 81

  1. Calculate Sum:

Sum = 60 x 81 = 4860

17
Q

Rules for Sum And Average of Consecutive Integers:

A

GMAT tests these so REMEMBER:

  1. Sum of terms in set of consecutive integers with an odd number of terms, is always a multiple of the number of terms.

E.g. 1, 2, 3 sum = 6 which is a multiple of number of terms, 3.

  1. Average of odd number of consecutive integers will always be an integer (equal to median of set).
  2. Sum of integers in consecutive sets with even number of terms will never be a multiple of the number of terms.

E.g. 1, 2, 3, 4 sum = 10 which is not a multiple of number of terms, 4.

  1. Average of even number of consecutive integers will never be an integer.

E.g. 1, 2, 3, 4
Average = 2+3/2 = 5/2 = 2.5

18
Q

How to Calculate Sum of Part of Integers in Set:

A

E.g. If the sum of the last 3 integers in a set of 6 consecutive integers is 624, what is the sum of the first 3 integers of the set?

Two ways to solve problems like that:

  1. Sum Formula:
S = A x #
624 = A x 3
208 = A

That’s the second to last term in the set of 6 because in a set of 3 the middle term (median) is the average.

So, set is:

204, 205, 206, 207, 208, 209

Sum of first 3 integers: 615.

  1. Think of 6 consecutive integers always as:

n, n+1, n+2, n+3, n+4, n+5

Then:
n+3 + n+4 + n+5 = 624
3n + 12 = 624
n = 204

Then you can set up your sequence and make calculations:

204, 205, 206, 207, 208, 209

19
Q

Formula for Middle Term of Series:

A

In a consecutive series of positive integers that has an odd number of terms (n) this is the formula for the middle term:

n + 1/2

and since in a consecutive series of integers with an odd number of terms the middle term is the average, n+1/2 is also the average of the series.

20
Q

Remainders in Consecutive Sets:

A

By definition, if you divide the numbers in a consecutive set of 7 numbers by 7 you will get all remainders from 0 to 6. It doesn’t matter where the set begins, with a number that is a multiple of 7 itself (which would leave remainder of 0) or somewhere else, by the time you have gone through all 7 integers in the set and divided them by 7 you will get all remainders in the set of integers from 0-6.

E.g. Let’s say our set of numbers begins with 11:

Set: 11, 12, 13, 14, 15, 16, 17

Remainders: 
11/7 = 4
12/7 = 5
13/7 = 6
14/7 = 0 
15/7 = 1
16/7 = 2
17/7 = 3

This applies to other numbers too, i.e. set of 8 consecutive numbers and each divided by 8 will give you a set of remainders of all numbers from 0 - 7.

21
Q

Products of Consecutive Integers and Divisibility:

A

Since there is always a multiple of three in a set of three consecutive integers, the product of the three consecutive integers will always be divisible by 3. There will also always be one multiple of 2 in any set of three consecutive integers. Therefore, the product of three consecutive integers will always divisible by 3!, so 6.

3x2x1 = 6

This rule applies to any number of consecutive integers:

Product of k consecutive integers is always divisible by k!

22
Q

Sum of Consecutive Integers and Divisibility:

A
  1. In any set of consecutive integers with ODD number of items, sum of all integers is always a multiple of number of items because:

Sum = Average x #
so:
Average = Sum/#

And you know that in a set of consecutive integers with odd number of items the average is always an integer and the median of the set.

    1. In any set of consecutive integers with EVEN number of items, sum of all integers is never a multiple of number of items because again:

Sum = Average x #
so:
Average = Sum/#

And you know that for an even number of integers, the average is never an integer, so, the sum can’t be a multiple of the # of items.

E.g. {1, 2, 3, 4}
Average = Sum/#
A = 10/4

10 is not a multiple of 4.

So REMEMBER this:

If:

Sum of k integers/k = Integer

it means that:

Average of k = integer

and if the average of consecutive set is an integer, then k MUST be odd.

23
Q

Consecutive Integers and Divisibility:

A

REMEMBER: In any set of consecutive integers with one even integer, the product of the set will always be divisible by 2. If there are two even integers in the set (like in set {2, 3, 4,}) the product will be divisible by 4 because 2 is a factor of the product twice. 2x2 = 4 is a factor of the product of the series.

REMEMBER that on GMAT consecutive integers can be somewhat hidden in expressions like this:

Is x(x+1)(x+2) divisible by 4? Essentially they are asking if the product of the set of consecutive integers is divisible by 4.

24
Q

Scheduling Problems:

A

In scheduling problems consider the extreme possibilities. E.g. if questions asks how many days a warranty lasts. Look at the info given and consider both extremes, shortest days possible and longest days possible, to get to the solution.

25
Q

Average and Mean in Multiples:

A

In a group of multiples of an integer the average and mean are also the same because the grouped in pairs there as well.

E.g. in sequence of first ten multiples of 5 the pairs are 5 and 50, 10 and 45, 15 and 40 etc. Because there are a perfect set of pairs the average and median are the same in this sequence, 27.5.

Average: 5 + 50/2 = 55/2 = 27.5
Median: because it’s a sequence with even amount of integers the median is the average of the middle two numbers: 25+30/2 = 55/2 = 27.5