WEEK ELEVEN                
TOPIC: Mean Deviation, Variance and standard Deviation of Grouped Data use in solving practical problems related to real life situations

 Mean Deviation of Grouped Data
Example 1
The speeds of 40 cars in a certain road are tabulated as follows:

Speed (km/h)50 – 5455 – 5960 – 6465 – 6960 – 7475 – 8080 – 84
Frequency51015121062

For this distribution, calculate

  1. The mean
  2. The mean deviation

 Solution
The complete table of the distribution is shown below.

Class intervalMid – value (xm)
50 – 5452526013.1765.85
55 – 5957105708.1781.5
60 – 6462159303.1747.55
65 – 6967128041.8321.96
60 – 7472107206.8368.3
75 – 8077646211.8370.98
80 – 8482216416.8333.66
Total   

 

  1. Mean, =

    The mean is 65.2km/h to 1 d.p.

  2. Mean deviation =

    The mean deviation is 6.5km/h

     EVALUATION

  3. Calculate the mean and the mean deviation of the following:
    1. 8, 5, 12, 8, 13, 4, 9, 5, 4, 7
    2. 9.25, 8.04, 12.08, 9.82, 10.05, 2.05, 8.25, 7.64, 7.02, 8.02

 Variance and Standard Deviation of a Grouped Data
Example 1
The table shows the time to the nearest hours of television watched by a group of students in a week.

Time1 – 56 – 1011 – 1516 – 2021 – 2526 – 3031 – 3536 – 40
Frequency2581014641

Calculate

  1. The mean
  2. The variance
  3. The standard deviation

 Solution
Let xm represents the mid-value (or class mark) of the interval.

  1. =

    Now subtract 19.8 from each value in the 2nd column to obtain the results in the 5th column. Then complete the other two columns as shown in the table.

  2. S2 =

    Variance = 64.8h to 3 s.f.

  3. S = = 8.047h

    Standard deviation is 8.05h to 3 s.f.

 Alternative method

Interval
1 – 5326918
6 – 10854064320
11 – 151381041691352
16 – 2018101803243240
21 – 2523143225297406
26 – 302861687844704
31 – 3533413210894356
36 – 403813814441444
Total   

 EVALUATION

  1. Calculate to 1 d.p the mean and standard deviation of the following numbers:
    1. 5, 7, 12, 10, 5, 15, 14, 9, 7, 8
    2. 6.5, 8.5, 6.5, 8.4, 6.9, 2.5, 6.2, 5.5

 GENERAL EVALUATION

  1. The table bellows shows the age distributions of a group of people.
Age (yrs)20 – 2930 – 3940 – 4950 – 5960 – 6970 – 79
Frequency35101372

Calculate:

  1. The mean age
  2. The variance
  3. The standard deviation

 READING ASSIGNMENT
Essential Mathematics for Senior Secondary 1 pgs 237 – 248

 WEEKEND ASSIGNMENT

  1. The lowest temperatures of a city in Asia for 10 consecutive days are recorded as: – 5oC, – 6oC, -5oC, 4oC, 0oC, 1oC, 2oC, 3oC, 4oC, 7oC. Find the mean deviation. A. 3.9 B. 4.0 C. 3.6 D. 6.4

Use the table below to answer question 2 to 4
A dice is thrown 100 times. The results are recorded as shown in the following table

Score123456
Frequency151817211415

Calculate:

  1. The mean score A. 4.0 B. 3.5 C. 1.0 D. 5.6
  2. The variance A. 2.7 B. 3.7 C. 2.1 D. 1
  3. The standard deviation A. 4 B. 5.1 C. 1.6 D. 7
  4. Find the variance of x, 2x, 3x, 4x, 5x, 6x, 7x, 8x, 9x and 10x. A. 8.25x29x2 B. 10x2 7.25x2

 THEORY

  1. The shoe sizes of a group of people are as follows:
Shoe size5678910111213
Frequency3814162010531

For this distribution, calculate the mean deviation

  1. The table below show the age distributions of a group of people.
Age (yrs)20 – 2930 – 3940 – 4950 – 5960 -6970 – 79
Frequency35101372

Calculate (a) the mean age (b) the variance (c) the standard deviation

 
 
 
 
 
 

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