Goye Farms and Agro-Consults

10/01/2026

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26/10/2025

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14/09/2025

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07/09/2025

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29/04/2024

Question

The weight (in kg) of 50 contestants at a competition is as follows:

65 66 67 66 64 66 65 63 65 68 64 62 66 64 67 65 64 66 65 67 65 67 66 64 65 64 66 65 64 65 66 65 64 65 63 63 67 65 63 64 66 64 68 65 63 65 64 67 66 64.

(a) Construct a frequenct table for the discrete data.

(b) Calculate, correct to 2 decimal places, the:

(i) mean ;
(ii) standard deviation of the data.

Solution:
Method 1.

x f fx x² fx²
62 1 62 3844 3844
63 5 315 3969 19845
64 12 768 4096 49152
65 14 910 4225 59150
66 10 660 4356 43560
67 6 402 4489 26936
68 2 136 4624 9248

∑fx = 3250 ∑f = 50 ∑fx² = 211735

Mean = ∑fx/ ∑f
= 3253/50
= 65.06kg

S.D = √ (∑fx²/∑f) - (∑fx/ ∑f)²

= Sqrt((211735/50) - (3253/50)²
= Sqrt (4234.7 - 4232.8)
= Sqrt(1.8564)
= 1.36 (2 decimal places)

Method 2

D = x - A, where A is the assumed mean of 65

X (kg) f D D² fD fD²

62 1 - 3 9 -3 9
63 5 - 2 4 -10 20
64 12 - 1 1 -12 12
65 14 0 0 0 0
66 10 1 1 10 10
67 6 2 4 12 24
68 2 3 9 6 18

∑fD = 3, ∑fD² = 93

SD = Sqrt [ (∑fD²/ ∑f) - ( ∑fD/ ∑f)²]

= Sqrt((93/50) - (3/50)² )
= Sqrt(1.8564)
= 1.36 2 dp

Students are advised to master any of the two formulae.

This is a discrete data. The approach is similar to grouped data, except, the determination of CLASS BOUNDARIES and CLASS MARKS.

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Daily Highlight

25/04/2024

Question

Using the method of completing the square, solve, correct to 2 decimal places, (x−2)/4 = (x+2)/2x. (WAEC)

Solution:

Cross mutiply the equation
4(x+2)= 2x(x-2)
4x + 8 = 2x^2 - 4x

4x + 4x +8 - 2x² = 0
-2x² + 8x + 8 = 0
Multiply both sides by ( -)
2x² - 8x = 8
Divide by 2
x² - 4x = 4
Add the square of half of the coefficient of x

x² - 4x + (-4/2)² = 4 + (-4/2)²
(x - 2)^2 = 4 + 2^2
(x - 2)^2 = 8
Square root of both sides

x - 2 = ⁺₋ (√8 )
x = 2 + √8 or 2 - √8

x = 4.828 or - 0.828
x = 4.83 or - 0.83 ( 2 decimal places)


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25/04/2024

Question

Find the sum of the Arithmetic Progression (AP) 1, 3, 5,..., 101.

Solution:

Students should note the following:

1. The nth term of an AP is given by a + (n - 1)d
Where
a = first term,
d = common difference
n = number of terms

2. Sum of AP = n/2 (a + l), where l is the last term.
That means, last term, l = a + (n-1)d

3. Sum of nth term of an AP is given by
n/2[ 2a + (n - 1)d]

Now,
a = 1, d = 3 - 1=2 or 5 - 3 = 2, l = 101

101 = 1 + (n -1)x 2
101 = 1 + 2n - 2
101 = 2n - 1
102 = 2n
n = 102/2 = 51
There are 51 terms

Similarly, Sn = n/2(a + l)
Sn = 51/2 (1 + 101)
= 51/2 (102)
= 25.5 x 102
= 2601. Answer


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25/04/2024

Question
A tree and a flagpole are on the same horizontal ground. A bird on top of the tree observes the top and bottom of the flagpole below it at angle of 45° and 60° respectively. If the tree is 10.65m high, calculate, correct to 3 significant figures, the height of the flagpole

Solution:

The students should take note of the following :

1. The tree is taller than the the flagpole, that is why the bird on it could observe the top and bottom of the flagpole.

2. The angles observed by the bird are of DEPRESSION

3. The tree and the flagpole are on the same horizontal ground, in opposite direction.

In the diagram above,

24/04/2024