Tuesday, October 23, 2007

Statistics - Rules and Tips!

1. Mean Average = total of quantities / number of quantities
2.
The median is the "middle" number in a group (when arranged in ascending or descending order) consisting of an odd number of numbers, and the average of the two middle numbers if there are an even number of numbers
3.
For a set of consecutive integers, the median is the the average of the first and the last integer
4.
Mode is the most frequently recurring number/numbers among the given set of numbers. It can be more than one
5.
Range is the difference between the largest number and smallest number is a set
6.
Calculation of Standard Deviation (SD):
  1. Find the mean, \scriptstyle\overline{x}, of the values.
  2. For each value xi calculate its deviation (\scriptstyle x_i - \overline{x}) from the mean.
  3. Calculate the squares of these deviations.
  4. Find the mean of the squared deviations. This quantity is the variance σ2.
  5. Take the square root of the variance.
7. Variance is the square of the standard deviation
8. SD does not change when the same constant is added or subtracted to all the members of the set
9.
If mean = maximum value it means that all values are equal and SD is 0
10.
A set of numbers with range of zero means that all of the numbers are the same, hence the dispersion of the numbers from its mean is zero
11.
For data with approximately the same mean, the greater the spread, the greater the SD.
12. SD is the square root of the average of the sum of square of the variation from the mean
13.
The more uneven members are dispersed around their arithmetic average, the more their SD
14.
You only need to know the difference between values and total number of values to compute SD
15.
If we know all the numbers of the list, there is a definite SD, regardless of what it is, we can compute it and get an answer – this is helpful for DS questions
16.
If the range is 0, then the SD must also be 0, because there is no variance
17.
The SD of any list is not dependent on the average, but on the deviation of the numbers from the average. So just by knowing that two lists having different averages doesn't say anything about their standard deviation - different averages can have the same SD
18.
The sum of the deviations of the elements from the mean must be 0
19.
Closer the more values to the MEAN, lower the SD
20. If
Range or SD of a list is 0, then the list will contain all identical elements
21.
Standard Deviation is also useful when comparing the spread of two separate data sets that have approximately the same mean. The data set with the smaller Standard Deviation has a narrower spread of measurements around the mean and therefore usually has comparatively fewer high or low values.
In general, the more widely spread the values are, the larger the
Standard Deviation is.
22. If you multiply all terms by x then SD =x times old SD and mean = x times old mean
23.
For comparing the SD for two sets any information about mean ,median,mode and range are insufficient unless you can determine the individual terms from the given data
24. Symmetric about the mean means that the shape of the distribution on the right and left side of the curve are mirror-images of each other
25.
For a given set of consecutive even numbers.. mean = median
26.
When you have a set of consecutive numbers (integers, evens, odds, multiples), the mean is equal to the median
27. Good links for Normal Distribution:

http://www-stat.stanford.edu/~naras/jsm/NormalDensity/NormalDensity.html

http://davidmlane.com/hyperstat/z_table.html


http://www.integratedlearning.net/gmat/sample.asp

Wednesday, October 03, 2007

Problem Solving - 29

IF s is the product of integers from 100 to 200, inclusive, and t is the product of integers from 100 to 201, inclusive, what is 1/s + 1/t in terms of t?

A. (201) ^2 / t
B. [(202) (201)]/t
C. 201/t
D. 202/t
E. [(202) (201)]/(t^2)

Answer: D

Given s = 100 * ... * 200
Given t = 100 * ... * 200 * 201
=> s = t/201
Hence 1/s + 1/t = 201/t + 1/t = 202/t

Problem Solving - 28

According to the directions on a can of frozen orange juice concentrate, 1 can of concentrate is to be mixed with 3 cans of water to make orange juice. How many 12-ounce cans of the concentrate are required to prepare 200 6-ounce servings of orange juice?

A. 25
B. 34
C. 60
D. 67
E. 100


Answer: A

4 cans of orange juice need 1 can of concentrate.
200 cans 6-ounce of orange juice need 50 cans 6-ounce concentrate.

so 12-ounce concentrate = 50/2 = 25 cans

OR

Let number of 12 ounce cans required = x
number of 12 ounce cans of water required = 3x

Total required servings of Orange juice = 200*6 ounce

Hence

x(12) + 3x(12) = 200*6
48x= 200*6
x=25

Hence A

Problem Solving - 27







In the addition table above, what is the value of m+n?

(A) -19
(B) 4
(C) 5
(D) 6
(E) 22


Answer: C

m = z + 4
n = y + e
m + n = z + 4 + y + e = (z+e) + (4+y) = 10 + (-5) = 5

Problem Solving - 26

n = 2*3*5*7*11*13/77k ...... If n is an integer and then which of the following could be the value of k?

(A) 22
(B) 26
(C) 35
(D) 54
(E) 60


Answer: B

n = 2*3*5*7*11*13/77k = 2*3*5*7*11*13/7*11*k = 2*3*5*13/k
=> k must be the product of any combination of product of these numbers.

From the given choices only 26 ------ (2*13) fits.

Problem Solving - 25

Two canoe riders must be selected from each of two groups of campers. One group consists of three men and one woman, and the other group consists of two women and one man. What is the probability that two men and two women will be selected?

(A) 1/6
(B) 1/4
(C) 2/7
(D) 1/3
(E) 1/2



Answer: E

1st Group: 3 Men, 1 Woman
2nd Group: 1 Man, 2 Women

Total number of ways of selection: 4C2 * 3C2 = 18

No of ways of selecting two men and two women:

1. 2 men and 0 woman from 1st group and 0 men and 2 women from 2nd group = 3C2 = 3
2. 1 man and 1 woman from 1st group and 1 man and 1 woman from 2nd group = 3C1* 2C1 = 6 Thus Total ways from the above = 3 + 6 = 9

Therefore Probability = 9/18 = 1/2