Exam Notes: Measures of Central Tendency, Position, and IQR

Measures of Central Tendency

Objectives

  • Recall and apply formulas for computing measures of central tendency (mean, median, and mode) for both grouped and ungrouped data.
  • Interpret and solve problems involving measures of position (percentiles and quartiles) in real-life and mathematical contexts for grouped and ungrouped data.
  • Analyze and compare data sets using computed measures to draw meaningful conclusions and support data-driven decisions.

Mean (Ungrouped Data)

  • To find the mean:
    • Add all values in the data set.
    • Divide by the number of values in the set.
  • Example:
    • Data set: 176, 178, 178, 189
    • Sum: 176+178+178+189=721176 + 178 + 178 + 189 = 721
    • Number of values: 4
    • Mean: 7214=180.25\frac{721}{4} = 180.25
    • The mean is 180.25 cm.

Median (Ungrouped Data)

  • To find the median:
    • Arrange all numbers in ascending order.
    • Determine the number in the middle.
  • If there is an even number of values, take the average of the two middle numbers.
  • Example:
    • Data set: 176, 178, 178, 189
    • Middle numbers: 178, 178
    • Median: 178+1782=3562=178\frac{178 + 178}{2} = \frac{356}{2} = 178
    • The median is 178.

Mode (Ungrouped Data)

  • To find the mode:
    • Arrange all numbers in ascending order.
    • The number that appears most often is the mode.
  • Example:
    • Data set: 176, 178, 178, 189
    • The mode is 178, as it occurs twice.

Question 1

  • Raymund is hosting a kiddie party with 7 kids aged 11 and 8 babies aged 1-3. Which measure of central tendency is appropriate to find the average age?
    • The best measure would be the mean.

Question 2

  • For the data set 12, 12, 13, 14, 14, which statement is true?
    • Mean: 12+12+13+14+145=655=13\frac{12 + 12 + 13 + 14 + 14}{5} = \frac{65}{5} = 13
    • Median: 13
    • Mode: 12 and 14
    • Therefore, Mean = Median.

Question 3

  • Ermie's quiz scores are 12, 10, 16, x, 13, and 9. What must be her score on the 4th quiz to get an average of 12?
    • Mean = 12, so 12+10+16+x+13+96=12\frac{12+10+16+x+13+9}{6} = 12
    • 60+x=7260 + x = 72
    • x=12x = 12

Question 4

  • What is the mode of set A data?
    • A = {43, 11, 94, 46, 43, 74}
    • The mode is 43.

Question 5

  • What is the median of the values in set B?
    • B = {4, 2, 9, 7, 5, 9, 3}
    • Arranged: {2, 3, 4, 5, 7, 9, 9}
    • The median is 5.

Mean (Grouped Data)

  • Formula: Mean=Σ(fx)Σf\text{Mean} = \frac{\Sigma(fx)}{\Sigma f}
    • Where:
      • ff is the frequency of each class.
      • xx is the class mark of each class (midpoint of the class interval).

Median (Grouped Data)

  • Formula: \text{Median} = lbmc + i \frac{(\frac{\Sigma f}{2} - <cf)}{fmc}
    • Where:
      • lbmclbmc is the lower boundary of the median class.
      • ff is the frequency of each class.
      • <cf is the cumulative frequency of the lower class next to the median class.
      • ii is the class interval.
    • The median class is the class with the smallest cumulative frequency greater than or equal to Σf2\frac{\Sigma f}{2}. The computed median must be within the median class.

Mode (Grouped Data)

  • Formula: Mode=lbmo+iD<em>1D</em>1+D2\text{Mode} = lbmo + i \frac{D<em>1}{D</em>1 + D_2}
    • Where:
      • lbmolbmo is the lower boundary of the modal class.
      • D1D_1 is the difference between the frequencies of the modal class and the next lower class.
      • D2D_2 is the difference between the frequencies of the modal class and the next upper class.
      • ii is the class interval.
    • The modal class is the class with the highest frequency.

Question 6 - 10 (Grouped Data Example)

  • Ages of 112 people on an island:
Xf
0-920
10-1921
20-2923
30-3916
40-4911
50-5910
60-697
70-793
80-891

Question 6

  • What is the class size of the given data?
    • The class size is 10 (e.g., 10 - 0 = 10).

Question 7

  • What is the modal class?
    • The modal class is 20-29, as it has the highest frequency (23).

Question 8

  • What is the value of the median class?
    • Total frequency is 112, so median is at 1122=56\frac{112}{2} = 56.
    • Cumulative frequencies: 20, 41, 64. The median class is 20-29.

Question 9

  • What is the ΣfXm\Sigma fXm?
    • This refers to the sum of the product of each class's frequency and midpoint. The answer choice is 3304 given from the slide.

Question 10

  • What is the mean of the data?
    • ΣfXmΣf=3304112=29.5\frac{\Sigma fXm}{\Sigma f} = \frac{3304}{112} = 29.5

Measures of Position and IQR

Percentiles

  • Divide the data set into one hundred equal parts.
  • There are ninety-nine percentiles denoted as P<em>1,P</em>2,P<em>3,…,P</em>99P<em>1, P</em>2, P<em>3, …, P</em>{99}.

Percentiles (Ungrouped Data)

  • Steps:
    1. Arrange the data set in ascending order.
    2. Locate the position of the percentile in the data set.
      • Pc=n×P100P_c = \frac{n \times P}{100}
      • Where:
        • nn = number of observations
        • PP = percentile you are finding. Ex: for 25th percentile, P=25P=25.
  • If PcP_c is not an integer, round it up to the next integer.
  • If PcP_c is an integer, calculate the mean of the observation in this position and the next (higher) position.

Interquartile Range (IQR)

  • Tells you the spread of the middle half of your distribution.
  • Is the difference between the upper and lower quartiles.
  • IQR=Q<em>3−Q</em>1IQR = Q<em>3 - Q</em>1
  • Quartiles divide the data into four equal parts (25% each).
    • Q1Q_1 (First Quartile): 25th percentile
    • Q2Q_2 (Second Quartile): 50th percentile (Median)
    • Q3Q_3 (Third Quartile): 75th percentile

Mendenhall and Sincich Method

  • Given a data set with nn number of values, first arrange the data in ascending order.
  • Odd Numbers:
    • L=Position of Q1=14(n+1)L = \text{Position of } Q_1 = \frac{1}{4}(n+1)
    • U=Position of Q3=34(n+1)U = \text{Position of } Q_3 = \frac{3}{4}(n+1)
  • Even Numbers:
    • L=Position of Q1=0.25(n)L = \text{Position of } Q_1 = 0.25(n)
    • U=Position of Q3=0.75(n)U = \text{Position of } Q_3 = 0.75(n)
  • IQR=Upper Quartile − Lower QuartileIQR = \text{Upper Quartile } - \text{ Lower Quartile}

Tukey's Method

  • To find Q1Q_1, get the median of the lower half of the data.

  • To find Q3Q_3, get the median of the upper half of the data.

  • IQR=Upper Quartile − Lower QuartileIQR = \text{Upper Quartile } - \text{ Lower Quartile}

  • Example:

    • Using Tukey's Method, find the first quartile and third quartile of the following set of values:
      • 5, 8, 9, 10, 11, 14, 15, 17, 18, 19, 21
      • Lower half: 5, 8, 9, 10, 11
        • Q1Q_1 = 9 (Median of lower half)
      • Upper half: 15, 17, 18, 19, 21
        • Q3Q_3 = 18 (Median of upper half)

Question 1

  • Angela scored in the 82nd percentile for aptitude in accounting. What percentage scored above?
    • 100% - 82% = 18%. Therefore, 18% scored above.

Question 2

  • Timothy scored 83/100 with a percentile rank of 72. Clayton scored 85/100 with a percentile rank of 70. Who performed better?
    • Timothy performed better because his percentile score is higher. This means he performed better than a larger percentage of his class than Clayton did relative to his. (C)

Question 3

  • Data: 10, 13, 13, 14, 15, 15, 16
    • Minimum: 10
    • Q1Q_1: 13
    • Median: 14
    • Q3Q_3: 15
    • Maximum: 16
    • IQR: 15 - 13 = 2

Question 4

  • Data: 5, 8, 8, 9, 10, 10, 11, 12, 13
    • Minimum: 5
    • Q1Q_1: 8
    • Median: 10
    • Q3Q_3: 12
    • Maximum: 13
    • IQR: 12 - 8 = 4

Question 5

  • Which measure divides the data set into one hundred equal parts?
    • Percentile (D)

Question 6

  • Which measure has exactly 25% of the distribution below it?
    • First Quartile (A)

Question 7

  • Esmeralda's NCAE score is at the 99th percentile. What does this mean?
    • Esmeralda scored higher than the other 99% of the examinees. (D)