SPM Practice (Short Questions)

Posted on April 23, 2020 by user

Question 1:

A box contains 5 pink marbles and 21 yellow marbles. Sharon puts another 4 pink marbles and 1 yellow marble inside the box. A marble is chosen at random from the box.

What is the probability that a pink marble is chosen?

Solution:
Total number of pink marbles = 5 + 4 = 9
Total number of yellow marbles = 21 + 1 = 22
Total number of marbles = 9 + 22 = 31

∴ P (pink marble) = \frac{9}{31}

Question 2:
In a group of 80 prefects, 25 are girls. Then 10 boys leave the group.
If a prefect is chosen at random from the group, state the probability that the prefect chosen is a boy.

Solution:
Number of boys = 80 – 25 = 55

After 10 boys left,
Number of boys = 55 – 10 = 45

Total number of prefects = 25 + 45 = 70

∴ P (b o y) = \frac{45}{70} = \frac{9}{14}

Question 3:
60 people have taken part in a singing contest. If a person is chosen at random from all the contestants, the probability of getting a male contestant is

\frac{8}{15}

. If there are 12 males and 3 females who did not qualify to the second round, find the probability that a male is chosen from the second round contestants.

Solution:
Number of male contestants = 60 ×

\frac{8}{15}

= 32

After 12 males and 3 females left,
Total number of contestants left = 60 – 15 = 45

Number of male contestant = 32 – 12 = 20

∴ P (male contestant) = \frac{20}{45} = \frac{4}{9}

7.3 Probability of an Event

Posted on April 23, 2020 by user

(A) Probability of an Event

1. The probability of an event A, P(A) is given by

2. If P(A) = 0, then the event A will certainly not occur.

3. If P(A) = 1, then the event A will certainly to occur.

Example 1:

Table below shows the distribution of a group of 80 pupils playing a game.

	Form Four	Form Five
Girls	28	16
Boys	12	24

A pupil is chosen at random from the group to start the game.

What is the probability that a boy from Form Five will be chosen?

Solution:

Let

A = Event that a boy from Form Five

S = Sample space

n(S) = 28 + 12 + 16+ 24 = 80

n(A) = 24

\begin{array}{l} P (A) = \frac{n (A)}{n (S)} \\ = \frac{24}{80} = \frac{3}{10} \end{array}

(B) Expected Number of Times an Event will Occur

If the probability of an event A and the number of trials are given, then the number of times event A occurs

= P(A) × Number of trials

Example 2:

In a football training session, the probability that Ahmad scores a goal in a trial is ⅝. In 40 trials are chosen randomly, how many times is Ahmad expected to score a goal?

Solution:

Number of times Ahmad is expected to score a goal

= ⅝ × 40

= 25

(C) Solving Problems

Example 3:

Kelvin has 30 white, blue and red handkerchiefs. If a handkerchief is picked at random, the probability of picking a white handkerchief is

\frac{2}{5} .

Calculate

(a) the number of white handkerchiefs.

(b) the probability of picking a blue handkerchief if 8 of the handkerchiefs are red in colour.

Solution:

Let

W = Event that a white handkerchief is picked.

B = Event that a blue handkerchief is picked.

R = Event that a red handkerchief is picked.

S = Sample space

(a)
n(S) = 30

\begin{array}{l} n (W) = P (W) \times n (S) \\ = \frac{2}{5} \times 30 = 12 \end{array}

(b)
Given n(R) = 8

n(B) = 30 – 12 – 8 = 10

\begin{array}{l} P (B) = \frac{n (B)}{n (S)} \\ = \frac{10}{30} = \frac{1}{3} \end{array}

7.1 Sample Space

Posted on April 23, 2020 by user

7.1 Sample Space

(A) Experiment

1. An experiment is a process or an action in making an observation to obtain the required results.

2. The result of the experiment is called the outcome.

(B) Listing all the possible outcomes of an experiment

We can list all the possible outcomes of an experiment by carrying out the experiment or by reasoning.

Example 1:

Six cards as shown in the diagram above are placed in a box. A card is drawn at random from the box. List all the possible outcomes.

Solution:

All the possible outcomes are O, R, A, N, G, E.

(C) Sample space of an experiment

1. A sample space is the set of all the possible outcomes of an experiment.

2. The letter S is used to represent the sample space and all the possible outcomes are written in brackets, { }.

Example 2:

A letter is chosen from the word ‘GARDEN’.

(a) List all the possible outcomes.

(b) Write the sample space, S, using set notation.

Solution:

(a) The possible outcomes are G, A, R, D, E and N

(b) Sample space, S = {G, A, R, D, E, N }

7.2 Events

Posted on April 23, 2020 by user

7.2 Events

(A) Elements of a Sample Space which Satisfy Given Condition

When a specific condition is given, we can list the elements of a sample space which satisfy the given condition.

Example 1:

A two-digit number which is not more than 25 is chosen at random. List the elements of the sample space which satisfy each of the following conditions.

(a) A perfect square is chosen.

(b) A prime number is chosen.

Solution:

Sample space

= {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}

(a)

A perfect square is chosen

= {16, 25}

(b)

A prime number is chosen

= {11, 13, 17, 19, 23}

(B) Events for a Sample Space

An event is a set of outcomes which satisfy a specific condition.

Event is a subset is a sample space.

Example 2:

A coin and a die are thrown simultaneously. The events P and Q are defined as follows.

P = Event of obtaining a ‘heads’ from the coin and an odd number from the die.

Q = Event of obtaining a ‘tails’ from the coin and a number more than 2 from the die.

(a) List the sample space, S.

(b) List the elements of

(i) The event P,

(ii) The event Q.

Solution:

(a)

Sample space, S

= { (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6) }

(b)(i)

P = { (H, 1), (H, 3), (H, 5) }

← (Set of outcomes of Event P: obtaining a ‘heads’ from the coin and an odd number from the die.)

(b)(ii)

Q = { (T, 3), (T, 4), (T, 5), (T, 6) }

← (Set of outcomes of Event Q: obtaining a ‘tails’ from the coin and a number more than 2 from the die.)

6.5 Cumulative Frequency

Posted on April 23, 2020 by user

6.5 Cumulative Frequency

Cumulative Frequency of a data or a class interval in a frequency table is obtained by determining the sum of its frequency with the total frequencies of all its previous data or class interval.

(A) Ogive

Ogive is a cumulative frequency graph which is obtained by plotting the cumulative frequency against the upper boundaries of each class.

Example:

The data below shows the number of books read by a group of 60 students in a year.

Books	Frequency
6-10	3
11-15	7
16-20	11
21-25	16
26-30	11
31-35	8
36-40	4

(a) Construct a cumulative frequency table for the given data.

(b) By using the scales of 2 cm to 5 books on the horizontal axis and 2 cm to 10 students on the vertical axis, draw an ogive for the data.

Solution:

(a)

Add a class with frequency 0 before the first class.
Find the upper boundary of each class interval.

Books	Frequency	Cumulative Frequency	Upper Boundary
1-5	0	0	5.5
6-10	3	0 + 3 = 3	10.5
11-15	7	3 + 7 = 10	15.5
16-20	11	10 + 11 = 21	20.5
21-25	16	21 + 16 = 37	25.5
26-30	11	37 + 11 = 48	30.5
31-35	8	48 + 8 = 56	35.5
36-40	4	56 + 4 = 60	40.5

(b)

6.7 SPM Practice (Long Questions)

Posted on April 23, 2020 by user

Question 3:

The data below shows the mass, in kg, of 50 students.

(a) Copy and complete the table below based on the data above.

(b) Based on the completed table above,

(i) State the size of the class interval used in the table.

(ii) Calculate the estimated mean of the mass of the students.

For this part of the question, use graph paper.
(c) By using a scale of 2 cm to 5 kg on the horizontal axis and 2 cm to 2 students on the vertical axis, draw a frequency polygon for the data.

Solution:
(a)

(b)(i)

Size of class interval

= upper boundary – lower boundary

= 44.5 – 39.5

= 5

(b)(ii)

\begin{array}{l} Estimated mean = \frac{42 \times 4 + 47 \times 9 + 52 \times 7 + 57 \times 9 + 62 \times 18 + 67 \times 3}{50} \\ = \frac{2785}{50} = 55.7 \end{array}

(c)

6.1 Class Intervals

Posted on April 23, 2020 by user

6.1 Class Interval

1. Data that consist of the measurement of a quantity can be grouped into few classes and the range of each class in known as the class interval.

(A) Class Limits and Boundaries

Lower Limit and Upper Limit

2. For class interval, for example 30 – 39, the smaller value (30) is known as the lower limit while the larger value (39) is known as the upper limit.

Lower Boundary and Upper Boundary

3. The lower boundary of a class interval is the middle value between the lower limit of the class interval and the upper limitof the class before it.

4. The upper boundary of a class interval is the middle value between the upper limit of the class interval and the lower limitof the class after it.

Example:

20 – 29

30 – 39

40 – 49

\begin{array}{l} Lower boundary of the class 30 - 39 \\ = \frac{29 + 30}{2} = 29.5 \\ Upper boundary of the class 30 - 39 \\ = \frac{39 + 40}{2} = 39.5 \end{array}

(B) Class size

5. The class size is the difference between the upper boundary and the lower boundary of the class.

Example:

Size of class interval 30 – 39

= Upper boundary – Lower boundary

= 39.5 – 29.5

= 10

6.3 Histograms

Posted on April 23, 2020 by user

6.3 Histograms

(A) Draw a histogram based on the frequency table of a grouped data

1. A histogram is a graphical representation of a frequency distribution.

2. A histogram consists of vertical rectangular bars without any spacing between them.

3. Steps for drawing a histogram

(a) Determine the lower boundaries and upper boundaries for each class interval.

(b) Choose suitable scales for the horizontal axis (x-axis) to represent class interval and the vertical axis (y-axis) to represent frequency.

Example:

The following frequency table shows the radii, in cm, of different types of trees in a garden.

Radii (cm)	Frequency
2.0 – 2.4	7
2.5 – 2.9	5
3.0 – 3.4	10
3.5 – 3.9	2
4.0 – 4.4	6
4.5 – 4.9	4

Draw a histogram to represent the above information.

Solution:

Radii (cm)	Frequency	Lower boundary	Upper boundary
2.0 – 2.4	7	1.95	2.45
2.5 – 2.9	5	2.45	2.95
3.0 – 3.4	10	2.95	3.45
3.5 – 3.9	2	3.45	3.95
4.0 – 4.4	6	3.95	4.45
4.5 – 4.9	4	4.45	4.95

6.7 SPM Practice (Long Questions)

Posted on April 23, 2020 by user

Question 1:

The data below shows the age of 25 tourists who visited a tourist spot.

(a) Copy and complete the table below based on the data above.

(b) Based on the completed table above,

(i) State the modal class.

(ii) Calculate the mean age of the tourists.

(c) For this part of the question, use graph paper.

By using a scale of 2 cm to 5 years on the horizontal axis and 2 cm to 1 tourist on the vertical axis, draw a histogram for the data.

Solution:
(a)

Calculation of midpoint for (age 6 – 10) =

\frac{6 + 10}{2} = 8

(b)(i)
Modal class = age 16 – 20 (highest frequency)

(b)(ii)

\begin{array}{l} mean age = \frac{3 \times 2 + 8 \times 5 + 13 \times 3 + 18 \times 8 + 23 \times 3 + 28 \times 4}{25} \\ = 16.4 \end{array}

(c)

6.6 Measures of Dispersion (Part 1)

Posted on April 23, 2020 by user

6.6 Measures of Dispersion

(A) Determine the range of a set of data

1. For an ungrouped data,

Range = largest value – smallest value.

2. For a grouped data,

Range = midpoint of the last class – midpoint of the first class.

Example 1:

Determine the range of the following data.

(a) 720, 840, 610, 980, 900

(b)

Time (minutes)	1 – 6	7 – 12	13 – 18	19 – 24	25 – 30
Frequency	3	5	9	4	4

Solution:

(a)

Largest value of the data = 980

Smallest value of data = 610

Range = 980 – 610 = 370

(b)

Midpoint of the last class

= ½ (25 + 30) minutes

= 27.5 minute

Midpoint of the first class

= ½ (1 + 6) minutes

= 3.5 minute

Range = (27.5 – 3.5) minute = 24 minutes