MATH 1530

Unit 1 Describing Data

2.3 Measures of Central Tendency

Mean, Median, Mode, Midrange

Notation:

  \(\sum\) denotes the sum of a set of values

 \(x\) is the variable usually used to represent the individual data values

 \(n\) lower case n represents the number of data values in a sample

 \(N\) upper case N represents the number of data values in a population

Arithmetic Mean:

  Mean of a Sample (this is a statistic) \(\overline{x}=\frac{\sum x}{n}\)

  Mean of a Population (this is a parameter) \(\mu=\frac{\sum x}{N}\)

Median:

• The median is the middle value.
• Position of the Median \(P_{50}=\frac{n+1}{2}\)
• Advantage The median is not affected by an extreme value.

Mode:

• Mode the value that occurs most often
• Bimodal data containing 2 modes
• Multimodal data containing 3+ modes
• No Mode no value that occurs most often
• Advantage can be used with nominal data

MidRange:

• The mean of the max and min values in the original data set
• Midrange=\(\frac{\left(x_{\min }+x_{\max }\right)}{2}\)
• Advantage The midrange is very easy to compute.
• Disadvantage The midrange is sensitive to extremes so it is rarely used.

Round off Rule for Measures of Center:

 Carry one more decimal place than is present in the original set of values.

1. Calculate the measures of central tendency for the life expectancies from the given data. Verify using technology.
   
   

   Country	Health $ per capita	Obesity %	Life expectancy	Universal Healthcare
   Canada	5292	30.1	82	yes
   China	420	7.3	76	no
   Germany	5411	22.7	81	yes
   Italy	3258	23.7	84	yes
   Japan	3703	3.5	84	yes
   Mexico	677	27.6	77	no
   Norway	9522	24.8	82	yes
   Switzerland	9674	21	83	yes
   United Kingdom	3377	29.8	82	yes
   United States	9403	35	79	no

   
   
   1. Mean: 81
   2. Median: 82
   3. Position of the Median: \(\frac{n+1}{2}=\frac{10+1}{2}=5.5\)
   4. Mode: 82
   5. Midrange: \(\frac{76+84}{2}=80\)
2. Using the table in the problem above, calculate the measures of central tendency for the obesity percentages from the given data. Verify using technology.
   1. Mean: 22.55
   2. Median: 24.25
   3. Position of the Median: \(\frac{n+1}{2}=\frac{10+1}{2}=5.5\)
   4. Mode: none
   5. Midrange: \(\frac{3.5+35}{2}=19.25\)
3. What do you notice about the measures for the 2 sets of data?



4. Shanita made a note of the weights of 8 baskets of fruit, but she spilled coffee on the fruit and the last number got smudged. If the average weight of the baskets is 12.5 pounds, how much was the smudged number? In other words, how much did the 8th basket of fruit weigh?
   

   Sum of basket weights is \(8*12.5 = 100\), so basket 8 must be \(100 – 86.8 = 13.2\)



5. The mean weight of three dogs is 60 pounds. Which of the following is possible? (Answer yes or no.)
   • One of the dogs weighs 100 pounds. yes
   • None of the dogs weighs more than 60 pounds. yes
   • Each dog weighs less than 60 pounds. no
   • The dogs weigh 25, 55, and 110 pounds. no
   • The dogs weigh 24, 52, and 104 pounds. yes
   • Altogether the dogs weigh 180 pounds. yes
6. The mean can alternately be defined as the balancing point. Total distance below the mean = total distance above the mean. The mean of 2, 3, 6, and 9 is 5. Note that 3 + 2 = 1 + 4.
   
   
   
   1. Which of the lines represents the mean of the data points shown below.

      C

   
   
   2. Which of the lines represents the mean of the data points shown below.

      B

Shapes of Distributions

Symmetric (Normal):

 Mean and Median are the same as Mode







Skewed To The Left (negatively skewed):

 Mean and Median are to the LEFT of the Mode (Left Tail)



Skewed To The Right (positively skewed):

 Mean and Median are to the RIGHT of the Mode (Right Tail)








Bimodal:

 Two Modes



Uniform:

 All frequencies are the same



Weighted Means and Means of Grouped Data

7. Calculate the semester grade point average (GPA) for the student with the following grades:

   Course	Credit Hours	Grade	Point Value for Grade	Quality Points (Points x hours)
   ENGL 1020	3	C	2	6
   MATH 1530	3	A	4	12
   COMM 1010	3	B	3	9
   BIOL 1110	4	C	2	8
   COLL 1000	1	B+	3.5	3.5
   Total	14			38.5

   \(\frac{38.5}{14}=2.75\)

Calculating by Formula:


\(\overline{x}=\frac{\sum(f \cdot x)}{\sum f} \quad\quad \text{Use class midpoints for x}\)



8. Find the estimated mean of the following data:
   
   

   SECONDS	FREQUENCY	MIDPOINTS
   5.0 – 5.9	5	5.45
   6.0 – 6.9	8	6.45
   7.0 – 7.9	10	7.45
   8.0 – 8.9	7	8.45
   9.0 – 9.9	3	9.45

   
   
   1. What is the estimated mean of the data? 7.30

      By Formula: \(\overline{x}=\frac{\sum(f \cdot x)}{\sum f}=\frac{5(5.45)+8(6.45)+10(7.45)+7(8.45)+3(9.45)}{33}=\frac{240.85}{33} \approx 7.30\)

   2. What is the upper limit of the 4th class? 8.9
   3. What is the class width? 1.0
   4. What is the shape of the distribution? Normal
   5. What class holds the median number of seconds? 7.0-7.9
9. Below is a frequency distribution of the ages of riders for one ride on the Dahlonega Mine Train at Six Flags. Find the estimated mean of the following data:
   
   

   Age of Riders, in years	Number of Riders
   10 - 14	1
   15 - 19	2
   20 - 24	7
   25 - 29	15
   30 - 34	4

   
   
   1. What is the estimated mean of the data? 25.3
   2. What is the upper limit of the 1st class? 14
   3. What is the class width? 5
      • Using LCL \(15-10=5\)
      • Using UCL \(29-24=5\)
      • Using midpt \(32-27=5\)
   4. What is the shape of the distribution? Left tail
   5. What class holds the median number of seconds? Class 25-29

      Position of the median: \(\frac{n+1}{2}=\frac{29+1}{2}=\frac{30}{2}=15\). Add frequencies until you reach 15.