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
distributions with the formula (1 − (𝑘12)) , 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 away from its mean is no more than 𝑎12 .
Chebyshev polynomials.
Chebyshev Bias
His name has a variety of spellings, all derived from his original Russian-language name Пафну́тий Льво́вич 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 A+ A A- B+ B C+ C D E G Total SMK P 2 13 6 10 12 23 12 25 15 2 120 SMK Q 0 3 12 4 23 17 24 23 12 6 124 SMK R 1 5 5 4 16 14 26 16 19 7 113 SMK S 1 2 5 9 9 10 15 22 27 12 112a.
SMK P =𝟏𝟎𝟑𝟏𝟐𝟎× 𝟏𝟎𝟎 = 𝟖𝟓. 𝟖𝟑 SMK Q =𝟏𝟎𝟔𝟏𝟐𝟒× 𝟏𝟎𝟎 = 𝟖𝟓. 𝟒𝟖 SMK R =𝟏𝟏𝟑𝟖𝟕 × 𝟏𝟎𝟎 = 𝟕𝟔. 𝟗𝟗 SMK S =𝟏𝟏𝟐𝟕𝟑 × 𝟏𝟎𝟎 = 𝟔𝟓. 𝟏𝟖
SMK P is the best in performance with the maximum number of student (percentage) who pass the exam.
b.
SMK P
Score f fx fx2 0 2 0 0 1 13 13 13 2 6 12 24 3 10 30 90 4 12 48 192 5 23 115 575 6 12 72 432 7 25 175 1225 8 15 120 960 9 2 18 162 120 603 3673
SMK Q
Score f fx fx2 0 0 0 0 1 3 3 3 2 12 24 48 3 4 12 36 4 23 92 368 5 17 85 425 6 24 144 864 7 23 161 1127 8 12 96 768 9 6 54 486 124 671 4125 Mean =∑ 𝑓𝑥∑ 𝑓 =603120= 5.025 Standard deviation =√∑ 𝑓𝑥∑ 𝑓2− 𝑥2 = √3673120 − (5.025)2 = 2.315 Mean =∑ 𝑓𝑥∑ 𝑓 =671124= 5.411 Standard deviation =√∑ 𝑓𝑥∑ 𝑓2− 𝑥2 = √4125124 − (5.411)2 = 1.996
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.
SMK R
Score f fx fx2 0 1 0 0 1 5 5 5 2 5 10 20 3 4 12 36 4 16 64 256 5 14 70 350 6 26 156 936 7 16 112 784 8 19 152 1216 9 7 63 567 113 644 4170
SMK S
Score f fx fx2 0 1 0 0 1 2 2 2 2 5 10 20 3 9 27 81 4 9 36 144 5 10 50 250 6 15 90 540 7 22 154 1078 8 27 216 1728 9 12 108 972 112 693 4815 Mean =∑ 𝑓𝑥∑ 𝑓 =644113= 5.699 Standard deviation =√∑ 𝑓𝑥∑ 𝑓2− 𝑥2 = √4170113 − (5.699)2 = 2.103 Mean =∑ 𝑓𝑥∑ 𝑓 =112693= 6.188 Standard deviation =√∑ 𝑓𝑥∑ 𝑓2− 𝑥2 = √4815112 − (6.188)2 = 2.169
Part 2
a. 𝑃(𝑆𝑀𝐾𝑃) =1517 =59 b. i. 27C 5 = 80730 ii. 4C 2 × 23C3 = 10626 iii. 3C 3 ×24C2 = 276 c. i. 5P 5 = 120 ii. 2P 1 ×3P3 ×1P1 = 12 iii. = 3 × 2 × 2 × 1 × 1 = 12Part 3
a. 𝑛 = 27, 𝑝 =279 =13 , 𝑞 =23 i. 1827× 100 = 6623%ii. Standard Deviation = √𝑛𝑝𝑟 = √27 ×13×23 = 2.449 b. 𝑛 = 9, 𝑝 = 0.3 , 𝑞 = 0.7 i. 𝑃(𝑋 = 3) = 9C 3 (0.3)3 (0.7)6 = 0.2668 ii. 𝑃(𝑋 ≤ 2) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) 𝑃(𝑋 = 0) = 9C 0 (0.3)0 (0.7)9 = 0.0404 𝑃(𝑋 = 1) = 9C 1 (0.3)1 (0.7)8 = 0.1556 𝑃(𝑋 = 2) = 9C 2 (0.3)2 (0.7)7 = 0.2668
A
A+
A
A+
A
3 2 2 1 1A+
A
A
A
A+
2 3 2 1 1 School A+ A SMK P 2 13 SMK Q 0 3 SMK R 1 5 SMK S 1 2𝑃(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 =
𝑋−𝜇 𝜎𝑧 =
52−4615= 0.4
c. i. 𝑃(𝑋 ≥ 52) = 𝑃 (𝑧 ≥52−4615 ) = 𝑃(𝑧 ≥ 0.4) = 0.3446 ii. 𝑃(𝑋 < 30) = 𝑃 (𝑧 <30−4615 ) = 𝑃(𝑧 < −1.067) = 0.1430 d. 𝑃(30 ≤ 𝑋 ≤ 52) = 𝑃 (30−4615 ≤ 𝑧 ≤52−4615 ) = 𝑃(−1.067 ≤ 𝑧 ≤ 0.4) = 1 − 𝑃(𝑧 > 0.4) − 𝑃(𝑧 > 1.067) = 1 − 0.1430 − 0.3446 = 0.5124e. Let minimum score is m Total students = 469
Probability of top ten students = 46910 = 0.0213
f(z) 0.4 z 0 f(z) z 0 −1.067 f(z) z 0 −1.067 0.4 f(z) 𝑚 − 46 15 z 0 0.0213
𝑃(𝑋 > 𝑚) = 0.0213 𝑃 (𝑧 >𝑚 − 46 15 ) = 0.0213 𝑚 − 46 15 = 2.028 𝑚 = 76.42 f. 𝑃(𝑋 ≥ 40) = 𝑃 (𝑧 ≥40−4615 ) = 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 ≈ 307 g. The minimum score = 𝑛
𝑃(𝑋 > 𝑛) = 0.92 f(z) z 0 −𝑛 − 46 15 0.92 0.08
𝑃 (𝑧 >𝑛 − 46 15 ) = 0.92 −𝑛 − 46 15 = 1.406 𝑛 = 24.91
Part 5
a. Platinum , 𝐼 =6560× 100 = 108.33 Gold , 𝐼 =4845× 100 = 106.67 Silver , 𝐼 =4040× 100 = 100 Bronze , 𝐼 =3635× 100 = 102.86 b. Index Weightage 𝐼𝑊 Platinum 108.33 2 216.66 Gold 106.67 4 426.68 Silver 100 3 300 Bronze 102.86 1 102.86𝐼 =∑ 𝐼𝑊 ∑ 𝑊 =1046.210 = 104.62 c.
Medal Index Weightage
Platinum 108.33 × 110 100 = 119.16 2 Gold 106.67 × 100 100 = 106.67 4 Silver 100 × 95 100 = 95 3 Bronze 102.86 × 100 100 = 102.86 1 d. 𝐼 =∑ 𝐼𝑊∑ 𝑊 𝐼 =2(119.16) + 4(106.67) + 3(95) + 102.86 10 = 105.29 e. 𝐼 =𝐶𝐶16 14× 100 𝐶16 455× 100 = 105.29 𝐶16= 479.07