FINDING MEDIAN FOR UNGROUPED DATA

Question 1 :

Find the median of the given values : 47, 53, 62, 71, 83, 21, 43, 47, 41.

Solution :

Ascending order of the given data 

21, 41, 43, 47, 47, 53, 62, 71, 83

Number of values given  =  9 (odd)

  median  =  [(N + 1)/2]^th value

  Median  =  10/2  =  5th value.

Hence the median is 47.

Question 2 :

Find the Median of the given data: 36, 44, 86, 31, 37, 44, 86, 35, 60, 51

Solution :

Ascending order of the given data.

31, 35, 36, 37, 44, 44, 51, 60, 86, 86

Number of given observations  =  10

Median  =  (10/2) th value + [(10/2) + 1] th value

=  (5th value + 6th value)/2

=  (44 + 44)/2

=  44

Question 3 :

The median of observation 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 arranged in ascending order is 24. Find the values of x.

Solution :

The given data is in ascending order.

Number of observations of the given data is 10

Median is average of 5^th and 6^th observation.

[(x + 2) + (x + 4)]/2  =  24

2x + 6  =  48

2x  =  48 - 6

2x  =  42

x  =  21

Hence the value of x is 21.

Question 4 :

A researcher studying the behavior of mice has recorded the time (in seconds) taken by each mouse to locate its food by considering 13 different mice as 31, 33, 63, 33, 28, 29, 33, 27, 27, 34, 35, 28, 32. Find the median time that mice spent in searching its food.

Solution :

 31, 33, 63, 33, 28, 29, 33, 27, 27, 34, 35, 28, 32

Ascending order of given data is

27, 27, 28, 28, 29, 31, 32, 33, 33, 33, 34, 35, 63

Middle value is 7^th observation

Median  =  32

Question 5 :

Eight people work in a shop. They are paid hourly rates of,

$2, $15, $5, $4, $3, $4, $3, $3

a) Find

i) the mean     ii) the median     iii)  the mode

b) Which average would you use if you wanted to claim that staff were 

i) well paid     ii)  badly paid ?

c) What is the range ?

Solution :

$2, $15, $5, $4, $3, $4, $3, $3

i) Finding mean :

Total number of terms = 8

Sum of terms = 2 + 15 + 5 + 4 + 3 + 4 + 3 + 3

= 2 + 15 + 5 + 2(4) + 3(3)

= 22 + 8 + 9

= 39

Mean = 39/8

= 4.875

ii) Finding median :

Arranging the given rates from smallest to largest.

2, 3, 3, 3, 4, 4, 5, 15

Median term = [(n/2)^th term + [(n/2) + 1]^th term]/2

= (4^th + 5^th)/2

= (3 + 4)/2

= 7/2

= 3.5 is the median

iii) Finding the mode :

Mode is 3.

b) i) Well paid is $4.875

ii) badly paid is $3.

c) The range = largest value - smallest value

= 15 - 2

= 13 is the range.

Question 6 :

For the set of data consisting of

8, 8, 9, 10, 10

which statement is true?

a. mean = mode     b. median = mode

c. mean = median     d. mean < median

Solution :

Mean = [2(8) + 9 + 2(10)] / 5

= (16 + 9 + 20)/5

= 45/5

= 9

Median = 9

mode = 8 and 9 are the mode.

Mean and median are the same. So, option c is correct.

Question 7 :

A gardener buys 10 packets of seeds from two different companies. Each pack contains 20 seeds and he records the number of plants which grow from each pack.

a) Find the mean, median and mode for each company's seeds.

b) which company dose the mode suggest is best ?

c) Which company does the mean suggest is best ?

d) Find the range for each company's seeds.

Solution :

a) mean :

Company A :

= [5 + 6 + 8 + 7(20)] / (1 + 1 + 1 + 7)

= 159/10

= 15.9

Company B :

= [2(15) + 16 + 3(17) + 4(18)] / (2 + 1 + 3 + 4)

= (30 + 16 + 51 + 72)/10

= 169/10

= 16.9

Median :

Company A :

5, 6, 8, 20, 20, 20, 20, 20, 20, 20

Median = (20 + 20) /2

= 40/2

= 20 is the median

Company B :

15, 15, 16, 17, 17, 17, 18, 18, 18, 18

median = (17 + 17)/2

= 34/2

= 17 is the median

Mode :

For company A mode is 20

For company B mode is 18

b) Company A's mode is best.

c) Company B's mean is best.

d) Range of company A = 20 - 5 ==> 15

Range of company B = 18 - 15 ==> 3

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