PIE CHART EXERCISES

Pie chart exercises

1.Yusuf asked 24 students to choose their favourite sport..

He recorded the information in the table below so that he could draw a pie chart..

(a)  Complete the pie chart accurately to show this data.

Sports                                      volleyball              football                 hockey                  cricket

Number of students                  6                           9                           7                           2

Angle on pie chart                90⁰                             135⁰                          ?                                ?

(b) Which is the modal sport.

Working :

a.      For hockey = 360 x 7/24 = 105⁰

For cricket = 360 x 2/24 = 30⁰

Sports                                      volleyball      football           hockey              cricket

Number of students                  6                    9                      7                       2

Angle on pie chart                  90⁰                   135⁰                    105⁰                  30⁰

b.     Modal = Football.

PIE CHART EXAMPLES

Example 1 :

A country has three political parties , the Reds , the Blues  and the Greens. The pie chart shows the propotion of the total vote that each party received in an election.

(i)                  Find the value of x.

(ii)                What percentage of the votes did the Red party receive?

Working :

(i)                  X = 360 – 144 – 90  = 126

(ii)                40%

STATISTICAL DATA EXERCISE

STATICAL DATA EXERCISE :

Exercise 1 :

15 students estimated the area of the rectangle shown below.

The estimates , in square centimetres , were

 

                                                                  45 44 50 50 48 

24 50 46 43 50

48 20 45 49 47 

(i)                 Work out ,

a.       the mode                     b. the mean                              c. the median

Working :

(i)                 a. mode = 50

20 , 24 , 43 ,44, 45 , 45 , 46 , 47 , 48 , 48 ,  49 , 50 , 50 , 50 , 50.           

b.      Mean = 43.9

20 + 24 + 43 + 44 + 45 + 45 + 46 + 47 + 48 + 48 + 49 + 50 + 50 + 50 + 50.

                                                            15

           

c.       Median =  47

20 , 24 , 43 ,44, 45 , 45 , 46 , 47 , 48 , 48 ,  49 , 50 , 50 , 50 , 50

Exercise 2 :

The list show marks in an examination taken by a class of 10 students

65 , 51 , 35 , 34 , 12 , 51 , 50 , 75 , 48 , 39.

(i)                 Write down the mode.

(ii)              Work out the median.

(iii)            Calculate the mean.

Working :

(i)                 Mode = 51.

65 , 51 , 35 , 34 , 12 , 51 , 50 , 75 , 48 , 39.

(ii)              Median = 49

12 , 34 , 35 , 39 , 48 , 50 , 51 , 51 , 65 , 75.

48 + 50 = 49

     2

(iii)            Mean = 46

65 + 51 + 35 + 34 + 12 + 51 + 50 + 75 + 48  +39 = 460

460  = 46

        10

STATISTICAL DATA EXAMPLES

MEDIAN MODE AND MEAN EXAMPLES:

Example  :

1.Seven people were asked to guess the number of beans in a jar. Their guesses were 194 , 173 , 170 , 144 , 182 , 259 , 159.

(i)                  Find the median.

(ii)                Work out the mean.

Working :

(i)                  Median = 173

A. First  you have to arrange the numbers from smallest to the largest.

144 ,  159 , 170 , 173 , 182 , 194 ,  259.

B. And then you cancel the numbers one by one from each side.

144 , 159 , 170 , 173 , 182 , 194 , 259.

(ii)                Mean = 183

A.      First you have to add all the numbers given.

144 + 159 + 170 + 173 + 182 + 194 + 259 = 1281

B.      And then divide the answers that you got with the total numbers.

 1281  = 183

     7

Example  :

2.Naomi records the sizes of the 34 pairs of shoes that her shop sells in one day.

4  10  5  6  4   8  6   4   7  3  9  7  4

7   3   5  4  6   5  10  7  5  5  6  4  7

7  6   6   5  5   3   5   6

(i)                  Using the list above complete the frequency table :

Shoe size	3	4	5	6	7	8	9	10
frequency

(ii)                Calculate the mean of these shoe sizes.

(iii)              Calculate the percentage of all the pairs of shoes that are size seven(7).

Working :

(i)                  Using the list above complete the frequency table :

Shoe size	3	4	5	6	7	8	9	10
frequency	3	6	8	7	6	1	1	2

A.      First you have to arrange the numbers and then you count each numbers to find the frequency.

4  10  5  6  4   8  6   4   7  3  9  7  4                                3 , 3 , 3 , 4 , 4 , 4 , 4 , 4 , 4 , 5 , 5 , 5 , 5 , 5 , 5 , 5 , 5

7   3   5  4  6   5  10  7  5  5  6  4  7                                6 , 6 , 6 , 6 , 6 , 6 , 6 , 7 , 7 ,7 , 7 , 7 , 7 , 8 , 9 , 10 , 10

7  6   6   5  5   3   5   6

                            3 = 3  ,   4 = 6  ,  5 = 8  ,  6 =  7  ,  7 = 6  ,  8 = 1 , 9 = 1  ,  10 = 2

(ii)                Mean = 5.7

B.      Add all the numbers that are already arranged and divide it by the total numbers.

3 + 3 + 3 + 4 + 4 + 4 + 4 + 4 + 4 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 7 + 7 +7 + 7 + 7 + 7 + 8 + 9 + 10  + 10

                                                                                         34

(iii)              Percentage = 17.6 %

C.      Multiple the numbers of shoes size 7 by the total shoes pairs and divide by hundreds.

6 x 34  = 17.6 %

 100

MEAN , MODE, MEDIAN FORMULA.

STATISTICAL DATA

DEFINITION OF MEAN , MEDIAN , MODE AND RANGE :

MEDIAN IS THE MIDDLE VALUE

RANGE IS THE DIFFERENCES BETWEEN THE LOWEST AND THE HIGHEST VALUE

MODE IS THE MOST FREQUENT VALUE

MEAN IS THE AVERAGE

PROBABILITY EXERCISE

Probability Exercise :

1. What is the probability of getting a sum 9 from two throws of a dice?

Explanation :

                               In two throw of a dice , n (S) = (6x6)=36

                        Let E = event of getting a sum= {(3,6),(4,5),(5,4),(6,3)

                                   P(E) = n(e)/n(s) = 4/36 = 1/9

2. Two dice are rolled once. Calculate the probability that the sum of the numbers on the  two dice is 5
Possible outcomes (Sample Space) = {(1, 1), (1, 2),...............,(1, 6), (2, 1), (2, 2),................,(2, 6), (3, 1), (3, 2),...........,(3, 6), .............,(4, 1), (4, 2),..........,(4, 6), (5, 1), (5,2),...............,(5, 6), (6, 1), (6, 2),......................,(6, 6)}

Total possible outcomes = 36

Number of outcomes of the experiment that are favorable to the event that a sum of two events is 6

=> Favorable outcomes are: (1, 5), (2, 4), (3, 3), (4, 2) and (5, 1)

Number of favorable outcomes = 5.

Use, probability formula = Number offavorable outcomes / Total number of possible outcomes

= 536536

The probability of a sum of 6 is 536536.