Add Maths Project Kedah 2017

Part 1.1 Famous Statisticians : Pafutny Chebyshev. Pafnuty Chebyshev (1821-1894) is a Russian mathematician who is well known for Chebyshev’s Theorem, which extends the properties of normal distributions to other, non-normal 1

distributions with the formula (1 − (𝑘 2 )) , as long as the distribution’s z score‘s absolute value is less than or equal to k and the standard deviation is more than 1. The inequality was originally known as the Bienaymé-Chebyshev inequality after linguist Irenée-Jules Bienaymé, the author of the original theorem.

Career Pafutny Chebyshev was born on 16 May 1821 in Okatovo, Kaluga Region, Russia and Died on 8 December 1894 in St Petersburg, Russia. Over the course of his career he produced many notable papers, including papers on statistics, calculus, mechanics and algebra. In 1847, he was appointed to the University of St Petersburg after submitting a thesis titled On integration by means of logarithms. In 1850, he was promoted to extraordinary professor at St Petersburg. Pafutny Chebyshev is perhaps the most famous Russian mathematician and is considered the father of modern Russian mathematics.

Contributions to Mathematics Pafutny Chebyshev is probably most famous for the theorem that’s named after him. However, he did make several other notable mathematical contributions, including: 

The Chebyshev inequality (not to be confused with his Theorem) which states that if X is a random variable with standard deviation σ, then the probability that the outcome of X is no less than a\σ 1

away from its mean is no more than 𝑎2 . 

Chebyshev polynomials.



Chebyshev Bias

His name has a variety of spellings, all derived from his original Russian-language name Пафну́тий Льво́ вич Чебышёв. According to Princeton University, his name is alternatively spelled Chebychev, Chebyshov, Tchebycheff or Tschebyscheff (the latter two are French and German transcriptions). Fun fact: The moon crater Crater Chebyschev and the asteroid 2010 Chebyshev are named after him.

Chebyschev’s crater on the moon.

Part 1.2 School SMK P SMK Q SMK R SMK S

A+ 2 0 1 1

A 13 3 5 2

A6 12 5 5

B+ 10 4 4 9

B 12 23 16 9

C+ 23 17 14 10

C 12 24 26 15

D 25 23 16 22

E 15 12 19 27

G 2 6 7 12

Total 120 124 113 112

a. 𝟏𝟎𝟑

SMK P = 𝟏𝟐𝟎 × 𝟏𝟎𝟎 = 𝟖𝟓. 𝟖𝟑 𝟏𝟎𝟔

SMK Q = 𝟏𝟐𝟒 × 𝟏𝟎𝟎 = 𝟖𝟓. 𝟒𝟖 SMK R = SMK S =

𝟖𝟕 × 𝟏𝟎𝟎 = 𝟕𝟔. 𝟗𝟗 𝟏𝟏𝟑 𝟕𝟑 × 𝟏𝟎𝟎 = 𝟔𝟓. 𝟏𝟖 𝟏𝟏𝟐

SMK P is the best in performance with the maximum number of student (percentage) who pass the exam. b.

SMK P

SMK Q

Score 0 1 2 3 4 5 6 7 8 9

f 2 13 6 10 12 23 12 25 15 2 120

∑ 𝑓𝑥

2

fx 0 13 12 30 48 115 72 175 120 18 603

fx 0 13 24 90 192 575 432 1225 960 162 3673

603

∑ 𝑓𝑥 2 ∑𝑓

=√

fx2 0 3 48 36 368 425 864 1127 768 486 4125

fx 0 3 24 12 92 85 144 161 96 54 671

671

Mean = ∑ 𝑓 = 124 = 5.411 −𝑥

3673 − 120

= 2.315

f 0 3 12 4 23 17 24 23 12 6 124

∑ 𝑓𝑥

Mean = ∑ 𝑓 = 120 = 5.025 Standard deviation =√

Score 0 1 2 3 4 5 6 7 8 9

2

(5.025)2

Standard deviation =√

∑ 𝑓𝑥 2 ∑𝑓

4125

−𝑥

2

= √ 124 − (5.411)2 = 1.996

SMK S

SMK R Score 0 1 2 3 4 5 6 7 8 9

f 1 5 5 4 16 14 26 16 19 7 113 ∑ 𝑓𝑥

2

fx 0 5 10 12 64 70 156 112 152 63 644

fx 0 5 20 36 256 350 936 784 1216 567 4170

644

∑ 𝑓𝑥 2 ∑𝑓

4170

f 1 2 5 9 9 10 15 22 27 12 112 ∑ 𝑓𝑥

Mean = ∑ 𝑓 = 113 = 5.699 Standard deviation =√

Score 0 1 2 3 4 5 6 7 8 9

fx2 0 2 20 81 144 250 540 1078 1728 972 4815

fx 0 2 10 27 36 50 90 154 216 108 693

693

Mean = ∑ 𝑓 = 112 = 6.188 −𝑥

2

Standard deviation =√

∑ 𝑓𝑥 2 ∑𝑓

= √ 113 − (5.699)2

=√

= 2.103

= 2.169

−𝑥

4815 − 112

2

(6.188)2

SMK Q is the most consistent in student achievement ( 𝜎 = 1.996) , the standard deviation measures how concentrated the data are around the mean and the more concentrated, the smaller the standard deviation.

c. The two students with the grade B will made a huge impact on the performance of the subjects in SMK R. The standard deviation from 2.10 become 2.11 and the percentage from 76.99 drop to 76.58.

Part 2 15

a. 𝑃(𝑆𝑀𝐾𝑃) = 17 =

5 9

b. C5 = 80730 4 C2 × 23C3 = 10626 3 C3 × 24C2 = 276

i. ii. iii.

27

i. ii.

5

c. P5 = 120 P1 × 3P3 × 1P1 = 12

2

2

3

2

1

1

A+

A

A

A

A+

3

2

2

1

1

A

A+

A

A+

A

iii.

=3×2×2×1×1 = 12

Part 3 9

1

2

a. 𝑛 = 27, 𝑝 = 27 = 3 , 𝑞 = 3 i. ii.

18 27

2

× 100 = 66 3 %

Standard Deviation = √𝑛𝑝𝑟 1

2

= √27 × 3 × 3 = 2.449

b. 𝑛 = 9, 𝑝 = 0.3 , 𝑞 = 0.7 i. 𝑃(𝑋 = 3) = 9C3 (0.3)3 (0.7)6 = 0.2668 ii. 𝑃(𝑋 ≤ 2) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2)

𝑃(𝑋 = 0) = 9C0 (0.3)0 (0.7)9 = 0.0404 𝑃(𝑋 = 1) = 9C1 (0.3)1 (0.7)8 = 0.1556 𝑃(𝑋 = 2) = 9C2 (0.3)2 (0.7)7 = 0.2668

School

A+

A

SMK P

2

13

SMK Q

0

3

SMK R

1

5

SMK S

1

2

𝑃(𝑋 ≤ 2) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) = 0.0404 + 0.1556 + 0.2668 = 0.4628

Part 4 a. 𝑋~𝑁(46, 225) mean = 46 , standard devition = √225 = 15 b.

𝑧= 𝑧=

𝑋−𝜇

𝜎 52−46 15

f(z)

= 0.4

c. i.

𝑃(𝑋 ≥ 52) = 𝑃 (𝑧 ≥

52−46 ) 15

= 𝑃(𝑧 ≥ 0.4) = 0.3446

z

0.4

0

f(z) ii.

𝑃(𝑋 < 30) = 𝑃 (𝑧 <

30−46 ) 15

= 𝑃(𝑧 < −1.067) = 0.1430

−1.067 30−46 15

d. 𝑃(30 ≤ 𝑋 ≤ 52) = 𝑃 (

≤𝑧≤

z

0

52−46 ) 15

f(z)

= 𝑃(−1.067 ≤ 𝑧 ≤ 0.4) = 1 − 𝑃(𝑧 > 0.4) − 𝑃(𝑧 > 1.067) = 1 − 0.1430 − 0.3446 = 0.5124

−1.067

0

z

0.4

f(z) e. Let minimum score is m Total students = 469

0.0213

10

Probability of top ten students = 469 = 0.0213

0

𝑚 − 46 15

z

𝑃(𝑋 > 𝑚) = 0.0213 𝑚 − 46 𝑃 (𝑧 > ) = 0.0213 15 𝑚 − 46 = 2.028 15 𝑚 = 76.42

f.

𝑃(𝑋 ≥ 40) = 𝑃 (𝑧 ≥

40−46 ) 15

= 𝑃(𝑧 ≥ −0.4) = 1 − 𝑃(𝑧 ≥ 0.4) = 1 − 0.3446 = 0.6554 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 = 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 × 𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 = 0.6554 × 469 = 307.38 f(z)

≈ 307 g. The minimum score = 𝑛 𝑃(𝑋 > 𝑛) = 0.92

0.92

0.08

𝑛 − 46 − 15

0

z

𝑃 (𝑧 >

−

𝑛 − 46 ) = 0.92 15

𝑛 − 46 = 1.406 15 𝑛 = 24.91

Part 5 a. 65

 Platinum , 𝐼 = 60 × 100 = 108.33 48

 Gold , 𝐼 = 45 × 100 = 106.67

40 × 100 = 100 40 36 = 35 × 100 = 102.86

 Silver , 𝐼 =  Bronze , 𝐼 b. Platinum Gold Silver Bronze

Index 108.33 106.67 100 102.86

Weightage 2 4 3 1

𝐼𝑊 216.66 426.68 300 102.86

𝐼=

∑ 𝐼𝑊 ∑𝑊

=

1046.2 10

= 104.62 c.

d. 𝐼 =

Medal

Index

Weightage

Platinum

108.33 × 110 = 119.16 100

2

Gold

106.67 × 100 = 106.67 100

4

Silver

100 × 95 = 95 100

3

Bronze

102.86 × 100 = 102.86 100

1

∑ 𝐼𝑊 ∑𝑊

2(119.16) + 4(106.67) + 3(95) + 102.86 10 = 105.29

𝐼=

e. 𝐼 =

𝐶16 𝐶14

× 100

𝐶16 × 100 = 105.29 455 𝐶16 = 479.07