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DATA ANALYSIS PROBLEMS
GRE QUANT
CATEGORY 2003 2004
FLIGHT PROBLEMS 20% 22.1%
BAGGAGE 18.3 21.8
CUSTOMER SERVICE 13.1 11.3
OVERSALES OF
SEATS
10.5 11.8
REFUND 10.1 8.1
FARE 6.4 6
RESERVATION 5.8 5.6
TOURS 3.3 2.3
SMOKING 3.2 2.9
ADVERTISING 1.2 1.1
CREDIT 1 0.8
SPECIAL
ACCOMODATION
0.9 0.9
OTHERS 6.2 5.3
TOTAL 100% 100%
TOTAL NO. OF
COMPLAINTS
22998 13278
Question 1
Approximately, how many complaints concerning credit were received by in 2003
1
100
× 22998 = 229.98 approximately 230
Question 2
By approximately, what % of the total number of complaints decreased from 2003 to 2004
Decrease in the number of compaints =
22998−13278
22998
× 100 % = 42.26
Approximately 42 %
Question 3
Which of the following is true
a) In each of the years 2003 and 2004 complaints about flight problems, baggage and customer service
together accounted for more than 50 % of all customer complaints received by the airline
In 2003, sum of flight problems, baggage and customer service = 20 + 18.3 +13.1 > 50
In 2004, sum = 22.1 + 21.8 +11.3 > 50
The above statement is true
b)Number of special passenger accomodation complaints was unchanged from 2003 to 2004
In 2003, no. of special passenger accomodation complaints =
0.9
100
× 22998
In 2004, complaints =
0.9
100
× 13278
The complaints decreased in 2004
The above statement is false
c) From 2003 to 2004, number of flight problem complaints increased by more than 2 %
In 2003, number of flight complaints =
20
100
× 22998
In 2004, no of flight complaints =
22.1
100
× 13278
Number of flight complaints decreased
Statement is false
Question 4
600 applicants were rated from 1 to 50 points, The ratings had a mean of 32.5 points and
a standard deviation of 7.1 points. How many standard deviations above or below the mean is a
rating of 48 points, 30 points, 20 points
48 − 32.5
7.1
= 2.2
48 is approximately 2 standard deviations above the mean
30 − 32.5
7.1
= −0.4
30 is approximately 0.4 standard deviations below the mean
20 − 32.5
7.1
= −1.8
20 is approximately 1.8 standard deviations below the mean
****************************
UCjmCXXIjd03JQad8-rUzs0Q
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Data analysis problems

  • 2. CATEGORY 2003 2004 FLIGHT PROBLEMS 20% 22.1% BAGGAGE 18.3 21.8 CUSTOMER SERVICE 13.1 11.3 OVERSALES OF SEATS 10.5 11.8 REFUND 10.1 8.1 FARE 6.4 6 RESERVATION 5.8 5.6 TOURS 3.3 2.3 SMOKING 3.2 2.9 ADVERTISING 1.2 1.1 CREDIT 1 0.8 SPECIAL ACCOMODATION 0.9 0.9 OTHERS 6.2 5.3 TOTAL 100% 100% TOTAL NO. OF COMPLAINTS 22998 13278
  • 3. Question 1 Approximately, how many complaints concerning credit were received by in 2003 1 100 × 22998 = 229.98 approximately 230
  • 4. Question 2 By approximately, what % of the total number of complaints decreased from 2003 to 2004 Decrease in the number of compaints = 22998−13278 22998 × 100 % = 42.26 Approximately 42 %
  • 5. Question 3 Which of the following is true a) In each of the years 2003 and 2004 complaints about flight problems, baggage and customer service together accounted for more than 50 % of all customer complaints received by the airline In 2003, sum of flight problems, baggage and customer service = 20 + 18.3 +13.1 > 50 In 2004, sum = 22.1 + 21.8 +11.3 > 50 The above statement is true
  • 6. b)Number of special passenger accomodation complaints was unchanged from 2003 to 2004 In 2003, no. of special passenger accomodation complaints = 0.9 100 × 22998 In 2004, complaints = 0.9 100 × 13278 The complaints decreased in 2004 The above statement is false
  • 7. c) From 2003 to 2004, number of flight problem complaints increased by more than 2 % In 2003, number of flight complaints = 20 100 × 22998 In 2004, no of flight complaints = 22.1 100 × 13278 Number of flight complaints decreased Statement is false
  • 8. Question 4 600 applicants were rated from 1 to 50 points, The ratings had a mean of 32.5 points and a standard deviation of 7.1 points. How many standard deviations above or below the mean is a rating of 48 points, 30 points, 20 points 48 − 32.5 7.1 = 2.2 48 is approximately 2 standard deviations above the mean
  • 9. 30 − 32.5 7.1 = −0.4 30 is approximately 0.4 standard deviations below the mean 20 − 32.5 7.1 = −1.8 20 is approximately 1.8 standard deviations below the mean ****************************
  • 10. UCjmCXXIjd03JQad8-rUzs0Q For more videos on GRE QUANT VISIT MY YOUTUBE CHANNEL CHANNEL ID: