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1.2 Standard Form (Sample Questions)

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Example 1:
Find the value of 7.3 × 103 + 3.2 × 104 and express your answer in standard form.

Solution:
7.3 × 103 + 3.2 × 104
= 7.3 × 103 + 3.2 × 101 × 103
= [7.3 + (3.2 × 101)] × 103
= 39.3 × 103
= 3.93 × 104


Example 2:
Find the value of 3.3 × 105 + 6400 and express your answer in standard form.

Solution:
= 3.3 × 105 + 6.4 × 103
= 3.3 × 101 × 101 × 103+ 6.4 × 103
= 330 × 103 + 6.4 × 103
= (330 + 6.4) × 103
= 336.4 × 103
= 3.364 × 105

1.2 Standard Forms

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1.2.1 Standard Form (Part 1)
1. Standard form is a way of writing very large numbers or very small numbers in the form of A × 10n, where  1 ≤ A < 10  and n is a positive or a negative integer.

Example 1:
Express each of the following numbers in standard form.
(a) 7244
(b) 32567
(c) 750000
(d) 0.65
(e) 0.0428
(f) 0.000369

Solution:
(a) 7244 = 7.244 × 1000 = 7.244 × 103

(b) 32567 = 3.2567 × 10000 = 3.2567 × 104

(c) 750000 = 7.5 × 100000 = 7.5 × 105

(d) 0.65
= 6.5 × 1 10
= 6.5 × 10-1

(e) 0.0428
= 4.28 × 1 100
= 4.28 × 10-2

(f) 0.000369
= 3.69 × 1 10000
= 3.69 × 10-4


Example 2:
Express each of the following numbers in standard form.
(a) 63.4
(b) 2738
(c) 23000
(d) 428000000
(e) 0.0063
(f) 0.000000038

Solution:
(a) 63.4 = 6.34 × 10

(b) 2738 = 2.738 × 1000 = 2.738 × 10³

(c) 23000 = 2.3 × 10000 = 2.3 × 104

(d) 428000000 = 4.28 × 100000000 = 4.28 × 108

(e) 0.0063
= 6.3 × 1 1000
= 6.3 × 10-3
 
(f) 0.000000038
= 3.8 × 1 100000000
= 3.8 × 10-8
 

1.1 Significant Figures

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1.1.1 Significant Figures (Part 1)
1. Significant figures are the relevant digits in an integer or a decimal number which has been rounded up to a value according to a degree of accuracy.

2. In rounding off positive numbers to a given number of significant figures, the significance of zero is shown as below.

(a) All non-zero digits in a number are significant figures (s. f.).
Example:
(i) 568 (3 s. f.)
(ii) 36.97 (4 s. f.)

(b) All zeroes between non-zero digits are significant.
Example:
(i) 7001 (4 s. f.)
(ii) 3.04 (3 s. f.)
(iii) 22.054 (5 s. f.)

(c) All zeroes that lie on the right of a non- zero digit in a decimal are significant.
Example:
(i) 0.70 (2 s. f.)
(ii) 4.500 (4 s. f.)
(iii) 3.00 (3 s. f.)

(d) Zeroes that lie on the left of a non-zero digit in a decimal are not significant.
Example:
(i) 0.05 (1 s. f.)
(ii) 0.0040 (2 s. f.)
(iii) 0.07040 (4 s. f.)

(e) Zeroes at the end of a whole number are to be considered as non significant unless stated otherwise.
Example
(i) 40 (1 s. f.)
(ii) 3670 (3 s. f.)
(iii) 704200 (4 s. f.)


Example 1:
Round off the following numbers correct to three significant figures.
(a) 246 = 246 (3 s. f.)
(b) 2463 = 2460 (3 s. f.)
(c) 24632 = 24600 (3 s. f.)
(d) 0.00745 = 0.00745 (3 s. f.)
(e) 0.007453 = 0.00745 (3 s. f.)
(f) 0.007455 = 0.00746 (3 s. f.)
(g) 0.007403 = 0.00740 (3 s. f.)

Example 2:
Round off each of the following numbers to the number of significant figures indicated in brackets.
(a) 3548 (2 s. f.)
(b) 0.5089 (3 s. f.)
(c) 33.028 (1 s. f.)
(d) 0.40055 (3 s. f.)
(e) 0.681 (2 s. f.)
(f) 38.97 (3 s. f.)

Solution:
(a) 3500 (2 s. f.)
(b) 0.509 (3 s. f.)
(c) 30 (1 s. f.)
(d) 0.401 (3 s. f.)
(e) 0.68 (2 s. f.)
(f) 39.0 (3 s. f.)

1.3 SPM Practice (Short Questions)

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Question 9:
3.17× 10 −8 −1.20× 10 −9 =

Solution:
3.17× 10 −8 −1.20× 10 −9 =3.17× 10 −8 −( 0.120× 10 1 × 10 −9 ) =3.17× 10 −8 −0.120× 10 −8 =( 3.17−0.120 )× 10 −8 =3.05× 10 −8


Question 10:
Express  0.096 ( 2× 10 3 ) 3  in standard form.

Solution:
0.096 ( 2× 10 3 ) 3 = 0.096 8× 10 9 =1.2× 10 −11


Question 11:
Express 0.000064−3.5× 10 −6  in standard form.

Solution:
0.000064−3.5× 10 −6 =6.4× 10 −5 −3.5× 10 −6 =6.4× 10 −5 −0.35× 10 −5 =( 6.4−0.35 )× 10 −5 =6.05× 10 −5


Question 12:
200.7× 10 11  is written as 2.007× 10 b  in the standard form. State the value of b.

Solution:
200.7× 10 11 =2.007× 10 2 × 10 11                   =2.007× 10 13                 b=13


Question 13:
Find the product of 0.1985 and 0.5.
Round off the answer correct to two significant figures.

Solution:
0.1985 × 0.5
= 0.09925
= 0.10 (two significant figures)

1.3 SPM Practice (Short Questions)

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Question 14:
The area of a rectangular public parking area is 7.2 km2 . Its width is 2400 m. The length, in m, of the parking area is

Solution:
Area = Length × Width Length × 2.400 km = 7.2 k m 2 Length= 7.2 2.4            =3 km            =3000 m            =3× 10 3  m


Question 15:
It is given that 30 solid metal cylinders each with a radius of 70 cm and a height of 300 cm, are melted to make 40 identical solid spheres.
Find the volume, in cm3 , of each solid sphere.

Solution:
Volume of 40 spheres = Volume of 30 cylinders =30× 22 7 × 70 2 ×300 =138 600 000  cm 3 Volume of 1 sphere = 138 600 000 40 =3 465 000  cm 3 =3.465× 10 6  cm 3 =3.47× 10 6  cm 3

8.2 Normal Distribution

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8.2 Normal Distribution

(A) Continuous Random Variable
Continuous random variable is a variable that can take any infinite value in a certain range.

(B) Normal Distribution
1.   A continuous random variable, X, is normally distributed if the graph of its probability function has the following properties.
 


· Its curve has a bell shape and it is symmetrical at the line x = µ.
· Its curve has a maximum value at x = µ.
· The area enclosed by the normal curve and the x-axis is 1.

2.  The notation of X being normally distributed with a mean, µ and a variance, σ2 is X ~ N (µ, σ2).


(C) Standard Normal Distribution
If a normal random variable, X, has a mean, µ = 0 and a standard deviation, σ = 1, then X follows a standard normal distribution, i.e. X ~ N (0, 1).
 

(D) Curve of a Standard Normal Distribution
1.  The curve of a standard normal distribution has the following properties.
 
  
· Its curve is symmetrical at the vertical line that passes through the mean, µ = 0 and has a variance, σ2 = 1.
· Its curve has a maximum value at Z = 0.
· The area enclosed by the standard normal curve and the z-axis is 1.


(E) Converting a Normal Distribution to Standard Normal Distribution
A normal distribution can be converted to the standard normal distribution using the following formula:
Z = x − μ σ
where,
Z = standard score or z - score
X = value of a normal random variable
µ = mean of a normal distribution
σ = standard deviation of a normal distribution

Long Questions (Question 1 & 2)

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Question 1:
In a school examination, 2 students out of 5 students failed Chemistry.
(a)    If 6 students are chosen at random, find the probability that not more than 2 students failed Chemistry.
(b)   If there are 200 Form 4 students in that school, find the mean and standard deviation of the number of students who failed Chemistry.
 
Solution:
(a)
X− Number of students who failed Chemistry. X~B( n,p ) X~B( 6,  2 5 ) P(X=r)= c n r . p r . q n−r P(X≤2) =P( X=0 )+P( X=1 )+P( X=2 ) = C 6 0 ( 2 5 ) 0 ( 3 5 ) 6 + C 6 1 ( 2 5 ) 1 ( 3 5 ) 5 + C 6 2 ( 2 5 ) 2 ( 3 5 ) 4 =0.0467+0.1866+0.3110 =0.5443

(b)
X~B( n,p ) X~B( 200,  2 5 ) Mean of X =np=200× 2 5 =80 Standard deviation of X = npq = 200× 2 5 × 3 5 = 48 =6.93



Question 2:
5% of the supply of mangoes received by a supermarket are rotten.
(a) If a sample of 12 mangoes is chosen at random, find the probability that at least two mangoes are rotten.
(b)  Find the minimum number of mangoes that have to be chosen so that the probability of obtaining at least one rotten mango is greater than 0.85.
 
Solution:
(a)
X ~ B (12, 0.05)
1 – P (X ≤ 1)
= 1 – [P (X = 0) + P (X = 1)]
= 1 – [  C 12 0 (0.05)0 (0.95)12 +   C 12 1 (0.05)1 (0.95)11]
= 1 – 0.8816
= 0.1184

(b)
P (X ≥ 1) > 0.85
1 – P (X = 0) > 0.85
P (X = 0) < 0.15
  C n 0 (0.05)0(0.95)n < 0.15
n lg 0.95 < lg 0.15
n > 36.98
n = 37

Therefore, the minimum number of mangoes that have to be chosen so that the probability of obtaining at least one rotten mango is greater than 0.85 is 37.


8.2c Probability Of An Event

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Example:

The masses of pears in a fruit stall are normally distributed with a mean of 220g and a variance of 100g. Find the probability that a pear that is picked at random has a mass

(a) of more than 230g.
(b) between 210g and 225g.

Hence, find the value of h such that 90% of the pears weigh more than h g.

Solution:

m= 220g
σ = √100 = 10g
Let X be the mass of a pear.

(a)

P( X>230 ) =P( Z> 230−220 10 )← Convert to standard normal distribution using Z= X−μ σ =P( Z>1 ) =0.1587

(b)

P( 210<X<225 ) =P( 210−220 10 <Z< 225−220 10 )← Convert to standard normal distribution using Z= X−μ σ =P( −1<Z<0.5 ) =1−P( Z>1 )−P( Z>0.5 ) =1−0.1587−0.3085 =0.5328

For 90% (probability = 0.9) of the pears weigh more than h g, 
P(X > h) = 0.9
P(X < h) = 1 – 0.9
                = 0.1
From the standard normal distribution table,
P(Z > 0.4602) = 0.1
P(Z < –0.4602) = 0.1

h−220 10 =−0.4602 h−220=−4.602 h=215.4

 

 

 

8.1a Probability Of An Event That Follows A Binomial Distribution

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8.1a Probability of an Event that follows a Binomial Distribution
 
1. For a Binomial Distribution, the probability of obtaining r numbers of successes out of n experiments is given by

P ( X = r ) = c n r . p r . q n − r

where
P = probability
X = discrete random variable
r =  number of success (0, 1, 2, 3, …, n)
n = number of trials
p = probability of success in an experiment (0 < p <1)
q = probability of failure in an experiment (q = 1 – p)



Example 1
Kelvin has taken 3 shots in a shooting practice. The probability that Kelvin strikes the target is 0.6. X represents the number of times kelvin strikes the target.

(a)   List the elements of the binomial discrete random variable X.

(b) Calculate the probability for the occurrence of each of the elements of X.

(c)  Hence, plot a graph to represent the binomial probability distribution of X.
 
Solution:
(a) X = Number of times Kelvin strikes the target
 X = {0, 1, 2, 3}
 
(b) X ~ B (n, p)
  X ~ B (3, 0.6)

P ( X = r ) = c n r . p r . q n − r ( i ) P ( X = 0 ) = C 3 0 ( 0.6 ) 0 ( 0.4 ) 3 ← Probability of failure = 1 − 0.6 = 0.4 = 0.064 ( i i ) P ( X = 1 ) = C 3 1 ( 0.6 ) 1 ( 0.4 ) 2 = 0.288 ( i i i ) P ( X = 2 ) = C 3 2 ( 0.6 ) 2 ( 0.4 ) 1 = 0.432 ( i v ) P ( X = 3 ) = C 3 3 ( 0.6 ) 3 ( 0.4 ) 0 = 0.216


(c)