GRE: Statistics

Measures of central tendency:- The main measure of central tendency are the arithmetic mean, mode, median. The average (arithmetic mean) of a group of number is defined as the sum of the values divided by the number of values.

AVERAGE VALUE  =  Sum of values ÷  Number of values

MEDIAN: When there are odd numbers of terms in a set, the median of a set is the middle of terms in a set, median of a set is the middle term.When all the terms in a set are listed in sequential order. When there are even number of terms in a set, the median of of a set is the average of two middle terms.

MODE:-  Mode is the number that appears most frequently in a list of numbers.

MEASURES OF DISPERSION:- The main measures of dispersion are the range, interquartile range and standard deviation.

First Quartile, Q is the median of all the numbers below the median. The second quartile, Q2 is the median of the entire data set. The third Quartile, Q3 is the median of all the numbers above the median. The interquartile range is the difference between Q3 and Q1.

The range and interquartile range are often displayed on a Box AND Whisker plot also called a Box Plot.

Problem 1) For the following Set : 2,2,3,4,5,5,6,21

Calculate Q1, Q2, Q3, interquartile range. Plot box and whisker plot for the following data?

Sol. Median = Mean of 4th and 5th term= (4+s / 2) = 4.5 = Q

Q1 is the median of 5,5,6, and 21 , Which is 5.5.

Interquartile range :-Q3 – Q1 = 5.5 -2.5     = 3

RANGE = 21-2 = 19

 

Problem 2) Which of the following set of data applies to this box and whisker plot?

 

 

 

 

 

 

 

 

a)-4, -4, -2, 0, 0, 5

b)-4, 1, 1, 3, 4, 4

c)-4, -4, -3, 1, 5

d)-5, 3, 4, 5

e)-4, -4, -2, -2, 0, 0, 0, 5

Sol. From this box and Whisker Plot, We can easily reckon that  Q1 = – 3, Q2 =-1  and Q3 = 0

Median of ” -4,-4,-2,0,0,5” is -1 , Median of “ -4,-4,-2,-2,0,0,0,5” is -1

Q1 for Set “-4,-4,-2,0,0,5”= Median of “-4,-4,-2”= 4

Q1 for set “-4,-4,-2,-2,0,0,0,5” = Median of “-4,-4,-2,-2” = (-4-2 / 2) = ( – 3 )

“-4,-4,-2,-2,0,0,0,5” is the only answer choice left. We can choose it without checking if you are confident in your previous work.

Standard Deviation:- Standard deviation is a measure of how spread out a set of numbers is.

For a set (n1 , n2 , n3 , _ _ _ _  n ), in Which x is the mean.

Standard deviation = \(\sqrt { \frac { { ({ n }_{ 1 }-x) }^{ 2 }+{ ({ n }_{ 2 }-x) }^{ 2 }———{ ({ n }_{ n }-x) }^{ 2 } }{ n } }\) 

 

Problem 3) Calculate the following standard deviation for the data set 0,7,8,10 and 10?

Ans) Mean = (0+7+8+10+10 / 5 ) = 7

Standard deviation =

\(\sqrt { \frac { { (7-0) }^{ 2 }+{ (7-7) }^{ 2 }+{ (7-8) }^{ 2 }+{ (7-10) }^{ 2 }+{ (7-10) }^{ 2 } }{ 5 } } \\ =\sqrt { \frac { 49+1+9+9 }{ 5 } } =\sqrt { \frac { 68 }{ 5 } } =3.7\)

 

Normal Distribution:- In the normal distributions, data points tend to cluster around the mean properties of normal distribution curve:-

1)The mean is at the center and the highest point on the curve and because the distribution is symmetrical about the mean, the mean is also the median and mode.

2)The standard deviation since it is a measure of dispersion determines the width of normal distributors curve. The greater the standard deviation, the wider and flatter the normal distribution curve is.

3)The probability of a data point falling within one standard deviation above the mean or below the mean is 34%

4)The probability of a data point falling between (Mean-d ) and (Mean-2d) is 13.5%.

5)The probability of a data point falling between (Mean+d) and (Mean+2d) is 13.5%

NOTE: The values shown in the diagram above for various key areas of the curve are approximate. You are likely to see slightly different value in GRE Questions.

 

Problem 4) A Food manufacturer produces energy bars that have a mean weight of  50 grams. If given days production is 10,000 energy bars, how many of those bars would be expected to weight between 49.0 and 49.5 grams?

(Assume that the weight are normally distributed)

Sol.

MEAN = 50 Gram

68% of the bar weight between 49.5 gm and 50.5 gm. Since 68% Corresponds to the area represented between (mean+d) and (mean-d). Therefore standard deviation= 0.5 gm

13.5% of these bars weigh between 49 and 49.5%.

No. of bars weighing between 49 and 49.5 =

\(=(\frac { 13.5 }{ 1000 } \times 10,000)={ 1,350 }^{ Ans }\)

 

Problem 5) How values among the 8000 homeowners of town X are normally distributed, with a standard deviation of  $11,000 and a mean of $ 90,000.

  Quantity A  The number of homeowners in Town X , Whose home value is greater than $ 112,000.

  Quantity B     300

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given

Sol.

 Number of homeowners in town X whose home value is greater than $ 112,000

=2.5% of  8000

= (2.5 / 100) × 8000      = 200

Therefore , Quantity B is Greater.

 

Problem 6) On a  particular test whose score is distributed normally, the 2nd percentile is 1720, while the 84th percentile if 1,990. What score round to the nearest 10, most closely corresponds to the 16th percentile?

Sol.

                   Mean  -2x (deviation)  = 1720

Mean + x(deviation)   = 1990

3x(deviation)   =  270

Deviation=90

Mean = 1720+2 (90)= 1900

Mean  –  deviation = 1900-90 =  \({ 1810 }^{ Ans }\)

 

Problem 7) If a set of data consist of only the first ten positive multiples of 5, What is the interquartile range of the set?

Sol. The first ten positive multiples of 5 are : 5,10,15,20,25,30,35,40,45,50.

Q1 (Median of [5,10,15,20,25] ) = 15

Q3 (Median of [30,35,40,45,50] ) = 40

INTERQUARTILE Range = \({ Q }_{ 3 }-{ Q }_{ 1 }=40-15={ (25) }^{ Ans }\)

 

Problem 8) Exam grades among the students in Ms. Hashman’s class are normally distributed, & the 50th percentile is equal to a score of 77.

Quantity A: The number of students who scored less than 80 on the exam.

Quantity B: The number of students who scored greater than 74 on the exam.

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol.  The normal distribution is symmetrical around the mean. For any symmetrical distribution, the mean equals the median (also known as the 50th percentile).

Thus the number of students who scored less than 3 points above the mean (77+3=80) must be the same as the number of students who scored greater than 3 points below the mean (77-3=74).

 

 

 

Problem 9) Jane scored in the 68th percentile on a test & John scored in the 32nd percentile.

Quantity A:    The proportion of the class that received a score less than John’s score.

Quantity B:    The proportion of a class that scored equal to or greater than Jane’s score.

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol. Percentiles define the proportion of a group that scores below a particular  benchmark. Since John scored in the 32nd percentile, by definition, 32% of the class scored worse than John.

Quantity A is equal to 32%.

Jane scored in the 68th percentile, So 68% of class scored worse than She did. Since 100-68=32, 32% of the class scored equal to or greater than Jane. Quantity B is also equal to 32%.

 

Problem 10) The lengths of a certain population of earthworms are normally distributed with a mean length of 30 cm & a standard deviation of 3 cm. One of the worm is picked at random.

Quantity A: The probability that the worm is between 24 & 30 centimeters, inclusive.

Quantity B: The probability that the worm is between 27 & 33 centimeters, inclusive.

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol.

Quantity A equals the area between (mean & Mean-2 (standard deviation)) of the distribution. Therefore, the probability that the worm is between 24 & 30 centimeters is about 48%.

Quantity B equals the area between ((mean+standard deviation) & (mean-standard deviation)) of the distribution. Therefore, the probability that the worm is between 27 & 33 centimeters is 68%.

Therefore, Quantity B is greater.

 

Problem 11) A species of insect has an average means of 5.2 grams & a standard deviation of 0.6 grams. The mass of the insects follows a normal distribution.

Quantity A: The percent of the insects that have a mass between 5.2 & 5.8 grams.

Quantity B: The percent of the insects that have a mass between 4.9 & 5.5 grams.

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol.

Mean=5.2 grams, Standard Deviation=0.6 grams

The percent of the insects that have a mass between 5.2 & 5.8 gm= 34% (approx.)

However, Quantity B will require some estimating.

Now that 4.9 is halfway between 4.6 & 5.2, while 5.5 is halfway between 5.2 & 5.8.

The area between 4.9 & 5.5 is under the bigger part of the bell curve in the centre. Since the area under the centre is greater than the area between (mean & (mean-standard deviation)).

Therefore, Quantity B is greater.

 

Problem: Among the set {1, 2, 3, 4, 7, 7, 10, 10, 11, 14, 19, 19, 23, 24, 25, 26}, what is the ratio of the largest item in the 2nd Quartile to the average value in the 4th Quartile?

Sol.

Median = Mean of 18th & 19th term =

\(=(\frac { 10+11 }{ 2 } )=10.5\)

First Quartile \({ Q }_{ 1 }\) = Median of all numbers below the median

= Median of {1 2 3 4 7 7 10 10}

= Mean of {4 7} =

\(=(\frac { 4+7 }{ 2 } )=5.5\)

Third Quartile \({ Q }_{ 3 }\) = Median of all numbers above the median =

= Median of {11 14 19 19 23 24 25 26}

= Median of {19 23}

\(=(\frac { 19+23 }{ 2 } )=21\)

Now, we have calculated \({ Q }_{ 1 },{ Q }_{ 2 }\quad and\quad { Q }_{ 3 }\).

Elements of first Quartile are {1 2 3 4}

Elements of Second Quartile are {7 7 10 10}

Elements of Third Quartile are {11 14 19 19}

Elements of Fourth Quartile are {23 24 25 26}

Ratio of largest item in the 2nd Quartile to the average value in the 4th Quartile =

\(=\frac { 10 }{ 24.5 } =\frac { 100 }{ 245 }\)

DEVELOPED

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