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

distributions with the formula (1 β (π 2 )) , as long as the distributionβs z scoreβs absolute value is less than or equal to k and the standard deviation is more than 1. The inequality was originally known as the BienaymΓ©-Chebyshev inequality after linguist IrenΓ©e-Jules BienaymΓ©, 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 thatβs named after him. However, he did make several other notable mathematical contributions, including: ο·

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

away from its mean is no more than π2 . ο·

Chebyshev polynomials.

ο·

Chebyshev Bias

His name has a variety of spellings, all derived from his original Russian-language name ΠΠ°ΡΠ½ΡΜΡΠΈΠΉ ΠΡΠ²ΠΎΜ Π²ΠΈΡ Π§Π΅Π±ΡΡΡΠ². 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.

Chebyschevβs 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. πππ

SMK P = πππ Γ πππ = ππ. ππ πππ

SMK Q = πππ Γ πππ = ππ. ππ SMK R = SMK S =

ππ Γ πππ = ππ. ππ πππ ππ Γ πππ = ππ. ππ πππ

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

β ππ₯

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

β ππ₯ 2 βπ

=β

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 = β π = 124 = 5.411 βπ₯

3673 β 120

= 2.315

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

β ππ₯

Mean = β π = 120 = 5.025 Standard deviation =β

Score 0 1 2 3 4 5 6 7 8 9

2

(5.025)2

Standard deviation =β

β ππ₯ 2 βπ

4125

βπ₯

2

= β 124 β (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 β ππ₯

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

β ππ₯ 2 βπ

4170

f 1 2 5 9 9 10 15 22 27 12 112 β ππ₯

Mean = β π = 113 = 5.699 Standard deviation =β

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 = β π = 112 = 6.188 βπ₯

2

Standard deviation =β

β ππ₯ 2 βπ

= β 113 β (5.699)2

=β

= 2.103

= 2.169

βπ₯

4815 β 112

2

(6.188)2

SMK Q is the most consistent in student achievement ( π = 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. π(πππΎπ) = 17 =

5 9

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

i. ii. iii.

27

i. ii.

5

c. P5 = 120 P1 Γ 3P3 Γ 1P1 = 12

2

2

3

2

1

1

A+

A

A

A

A+

3

2

2

1

1

A

A+

A

A+

A

iii.

=3Γ2Γ2Γ1Γ1 = 12

Part 3 9

1

2

a. π = 27, π = 27 = 3 , π = 3 i. ii.

18 27

2

Γ 100 = 66 3 %

Standard Deviation = βπππ 1

2

= β27 Γ 3 Γ 3 = 2.449

b. π = 9, π = 0.3 , π = 0.7 i. π(π = 3) = 9C3 (0.3)3 (0.7)6 = 0.2668 ii. π(π β€ 2) = π(π = 0) + π(π = 1) + π(π = 2)

π(π = 0) = 9C0 (0.3)0 (0.7)9 = 0.0404 π(π = 1) = 9C1 (0.3)1 (0.7)8 = 0.1556 π(π = 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

π(π β€ 2) = π(π = 0) + π(π = 1) + π(π = 2) = 0.0404 + 0.1556 + 0.2668 = 0.4628

Part 4 a. π~π(46, 225) mean = 46 , standard devition = β225 = 15 b.

π§= π§=

πβπ

π 52β46 15

f(z)

= 0.4

c. i.

π(π β₯ 52) = π (π§ β₯

52β46 ) 15

= π(π§ β₯ 0.4) = 0.3446

z

0.4

0

f(z) ii.

π(π < 30) = π (π§ <

30β46 ) 15

= π(π§ < β1.067) = 0.1430

β1.067 30β46 15

d. π(30 β€ π β€ 52) = π (

β€π§β€

z

0

52β46 ) 15

f(z)

= π(β1.067 β€ π§ β€ 0.4) = 1 β π(π§ > 0.4) β π(π§ > 1.067) = 1 β 0.1430 β 0.3446 = 0.5124

β1.067

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

π β 46 15

z

π(π > π) = 0.0213 π β 46 π (π§ > ) = 0.0213 15 π β 46 = 2.028 15 π = 76.42

f.

π(π β₯ 40) = π (π§ β₯

40β46 ) 15

= π(π§ β₯ β0.4) = 1 β π(π§ β₯ 0.4) = 1 β 0.3446 = 0.6554 ππ’ππππ ππ π π‘π’ππππ‘π = ππππππππππ‘π¦ Γ πππ‘ππ ππ’ππππ π π‘π’ππππ‘π = 0.6554 Γ 469 = 307.38 f(z)

β 307 g. The minimum score = π π(π > π) = 0.92

0.92

0.08

π β 46 β 15

0

z

π (π§ >

β

π β 46 ) = 0.92 15

π β 46 = 1.406 15 π = 24.91

Part 5 a. 65

ο Platinum , πΌ = 60 Γ 100 = 108.33 48

ο Gold , πΌ = 45 Γ 100 = 106.67

40 Γ 100 = 100 40 36 = 35 Γ 100 = 102.86

ο Silver , πΌ = ο Bronze , πΌ b. Platinum Gold Silver Bronze

Index 108.33 106.67 100 102.86

Weightage 2 4 3 1

πΌπ 216.66 426.68 300 102.86

πΌ=

β πΌπ βπ

=

1046.2 10

= 104.62 c.

d. πΌ =

Medal

Index

Weightage

Platinum

108.33 Γ 110 = 119.16 100

2

Gold

106.67 Γ 100 = 106.67 100

4

Silver

100 Γ 95 = 95 100

3

Bronze

102.86 Γ 100 = 102.86 100

1

β πΌπ βπ

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

πΌ=

e. πΌ =

πΆ16 πΆ14

Γ 100

πΆ16 Γ 100 = 105.29 455 πΆ16 = 479.07

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distributions with the formula (1 β (π 2 )) , as long as the distributionβs z scoreβs absolute value is less than or equal to k and the standard deviation is more than 1. The inequality was originally known as the BienaymΓ©-Chebyshev inequality after linguist IrenΓ©e-Jules BienaymΓ©, 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 thatβs named after him. However, he did make several other notable mathematical contributions, including: ο·

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

away from its mean is no more than π2 . ο·

Chebyshev polynomials.

ο·

Chebyshev Bias

His name has a variety of spellings, all derived from his original Russian-language name ΠΠ°ΡΠ½ΡΜΡΠΈΠΉ ΠΡΠ²ΠΎΜ Π²ΠΈΡ Π§Π΅Π±ΡΡΡΠ². 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.

Chebyschevβs 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. πππ

SMK P = πππ Γ πππ = ππ. ππ πππ

SMK Q = πππ Γ πππ = ππ. ππ SMK R = SMK S =

ππ Γ πππ = ππ. ππ πππ ππ Γ πππ = ππ. ππ πππ

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

β ππ₯

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

β ππ₯ 2 βπ

=β

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 = β π = 124 = 5.411 βπ₯

3673 β 120

= 2.315

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

β ππ₯

Mean = β π = 120 = 5.025 Standard deviation =β

Score 0 1 2 3 4 5 6 7 8 9

2

(5.025)2

Standard deviation =β

β ππ₯ 2 βπ

4125

βπ₯

2

= β 124 β (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 β ππ₯

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

β ππ₯ 2 βπ

4170

f 1 2 5 9 9 10 15 22 27 12 112 β ππ₯

Mean = β π = 113 = 5.699 Standard deviation =β

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 = β π = 112 = 6.188 βπ₯

2

Standard deviation =β

β ππ₯ 2 βπ

= β 113 β (5.699)2

=β

= 2.103

= 2.169

βπ₯

4815 β 112

2

(6.188)2

SMK Q is the most consistent in student achievement ( π = 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. π(πππΎπ) = 17 =

5 9

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

i. ii. iii.

27

i. ii.

5

c. P5 = 120 P1 Γ 3P3 Γ 1P1 = 12

2

2

3

2

1

1

A+

A

A

A

A+

3

2

2

1

1

A

A+

A

A+

A

iii.

=3Γ2Γ2Γ1Γ1 = 12

Part 3 9

1

2

a. π = 27, π = 27 = 3 , π = 3 i. ii.

18 27

2

Γ 100 = 66 3 %

Standard Deviation = βπππ 1

2

= β27 Γ 3 Γ 3 = 2.449

b. π = 9, π = 0.3 , π = 0.7 i. π(π = 3) = 9C3 (0.3)3 (0.7)6 = 0.2668 ii. π(π β€ 2) = π(π = 0) + π(π = 1) + π(π = 2)

π(π = 0) = 9C0 (0.3)0 (0.7)9 = 0.0404 π(π = 1) = 9C1 (0.3)1 (0.7)8 = 0.1556 π(π = 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

π(π β€ 2) = π(π = 0) + π(π = 1) + π(π = 2) = 0.0404 + 0.1556 + 0.2668 = 0.4628

Part 4 a. π~π(46, 225) mean = 46 , standard devition = β225 = 15 b.

π§= π§=

πβπ

π 52β46 15

f(z)

= 0.4

c. i.

π(π β₯ 52) = π (π§ β₯

52β46 ) 15

= π(π§ β₯ 0.4) = 0.3446

z

0.4

0

f(z) ii.

π(π < 30) = π (π§ <

30β46 ) 15

= π(π§ < β1.067) = 0.1430

β1.067 30β46 15

d. π(30 β€ π β€ 52) = π (

β€π§β€

z

0

52β46 ) 15

f(z)

= π(β1.067 β€ π§ β€ 0.4) = 1 β π(π§ > 0.4) β π(π§ > 1.067) = 1 β 0.1430 β 0.3446 = 0.5124

β1.067

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

π β 46 15

z

π(π > π) = 0.0213 π β 46 π (π§ > ) = 0.0213 15 π β 46 = 2.028 15 π = 76.42

f.

π(π β₯ 40) = π (π§ β₯

40β46 ) 15

= π(π§ β₯ β0.4) = 1 β π(π§ β₯ 0.4) = 1 β 0.3446 = 0.6554 ππ’ππππ ππ π π‘π’ππππ‘π = ππππππππππ‘π¦ Γ πππ‘ππ ππ’ππππ π π‘π’ππππ‘π = 0.6554 Γ 469 = 307.38 f(z)

β 307 g. The minimum score = π π(π > π) = 0.92

0.92

0.08

π β 46 β 15

0

z

π (π§ >

β

π β 46 ) = 0.92 15

π β 46 = 1.406 15 π = 24.91

Part 5 a. 65

ο Platinum , πΌ = 60 Γ 100 = 108.33 48

ο Gold , πΌ = 45 Γ 100 = 106.67

40 Γ 100 = 100 40 36 = 35 Γ 100 = 102.86

ο Silver , πΌ = ο Bronze , πΌ b. Platinum Gold Silver Bronze

Index 108.33 106.67 100 102.86

Weightage 2 4 3 1

πΌπ 216.66 426.68 300 102.86

πΌ=

β πΌπ βπ

=

1046.2 10

= 104.62 c.

d. πΌ =

Medal

Index

Weightage

Platinum

108.33 Γ 110 = 119.16 100

2

Gold

106.67 Γ 100 = 106.67 100

4

Silver

100 Γ 95 = 95 100

3

Bronze

102.86 Γ 100 = 102.86 100

1

β πΌπ βπ

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

πΌ=

e. πΌ =

πΆ16 πΆ14

Γ 100

πΆ16 Γ 100 = 105.29 455 πΆ16 = 479.07

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