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Genre: SoundtrackDate: 2008-2009Country: JapanAudio codec: MP3Quality: 320 kbsPlaytime: 3:51:31Character Song Collection 101. Nijiiro kyarachange! (3:34)02. Meikyuu butterfly (5:01)03. Happy Xmas (4:02)04. Ii koto arisou (3:49)05. Kimi no Birthday (3:23)06. Blue Moon (4:15)07. BLACK DIAMOND (3:46)08. Saikyou LOVE POWER (3:47)09. Heartful Song (4:46)10. Yume no tsubomi (1:27)Character Song Collection 201. Colorful heartbeat (3:55)02. Secret princess (4:00)03. Chiisana hoshi ~The Little Prince~ (4:34)04. Itsuka wa romance (5:15)05. Ookiku naare! (4:07)06. Hana tegami (4:25)07. Tsukiyo no marionette (3:57)08. Taiyou ga niau yo (3:16)09. Secret princess (3:59)OST01. Open Heart (1:42)02. Avant Title (0:14)03. Seiyo Gakuen no Asa (1:35)04. Subtitle (0:08)05. Cool & Spicy Girl!! (2:13)06. Hontou no Jibun (1:34)07. Kyou mo Ii Tenki (1:48)08. Hacha Mecha High Tension!? (1:37)09. Shinobiyoru Kage (1:49)10. X Chara Tanjou (1:47)11. Zettai Zetsumei! (1:36)12. Chara Nari! Amulet Heart (0:43)13. Chara Nari! Amulet Spade (0:43)14. Chara Nari! Amulet Clover (0:42)15. Battle (1:40)16. Kibou no Tamago Tachi (1:47)17. Gentle light (1:35)18. Eyecatch A (0:09)19. Eyecatch B (0:08)20. Jikai Yokoku (0:35)21. Hoshi wo Miagete (2:15)22. It’s A Shugo Chara World! (1:35)23. Watashi no Ouji-sama (1:36)24. Yamato Nadeshiko (1:36)25. Yamiyo wo Kakeru Kuro Neko (1:45)26. Tsukiyo no Violinist (2:24)27. Jounetsu wo Himete (1:46)28. Takuramu Mono Tachi (1:47)29. EASTER (1:34)30. Houkago (1:39)31. Tokimeku Kokoro (1:43)32. Futari no Kyori (1:48)33. Itsudemo Issho (1:36)34. Sunao na Kimochi de (1:46)OST201. Royal Garden ni Youkoso (1:52)02. Chiisana Ou-sama (1:39)03. King’s Chair (1:38)04. Ware wa Ou Nari (1:54)05. Nadeshiko no Mai (1:34)06. Konpeki no Yoru (1:46)07. Oh Dere Mademoiselle! (1:42)08. Bishitto Iu wa yo! (1:44)09. Hizashi Furu Machi (1:43)10.

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-3., 12.], [ -5., 18.]]) >>> 35 36 Numpy – array mathematics>>> A = np.array([[n+m*10 for n in range(5)] for m in range(5)]) >>> v1 = arange(0, 5) >>> A array([[ 0, 1, 2, 3, 4], [10, 11, 12, 13, 14], [20, 21, 22, 23, 24], [30, 31, 32, 33, 34], [40, 41, 42, 43, 44]]) >>> v1 array([0, 1, 2, 3, 4]) >>> np.dot(A,A) array([[ 300, 310, 320, 330, 340], [1300, 1360, 1420, 1480, 1540], [2300, 2410, 2520, 2630, 2740], [3300, 3460, 3620, 3780, 3940], [4300, 4510, 4720, 4930, 5140]]) >>> >>> np.dot(A,v1) array([ 30, 130, 230, 330, 430]) >>> np.dot(v1,v1) 30 Alternatively, we can cast the array objects to the type matrix. This changes the behavior of the standard arithmetic operators +, -, * to use matrix algebra. >>> M = np.matrix(A) >>> v = np.matrix(v1).T >>> v matrix([[0], [1], [2], [3], [4]]) >>> M*v matrix([[ 30], [130], [230], [330], [430]]) >>> v.T * v # inner product matrix([[30]]) # standard matrix algebra applies >>> v + M*v [131], [232], [333], [434]]) 36 37 Introduction to languageEnd

Latest Statistics MCQ Objective Questions Statistics Question 1: The mean of five observations is 4 and their variance is 5.2 . If three of these observations are 1, 2 and 6, then the other two observations are 4, 72, 105, 62, 94, 5Answer (Detailed Solution Below) Option 1 : 4, 7 Concept Used:Mean: \(\bar{x} = \frac{\sum x_i}{n}\)Variance: \(\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2\)Calculation:Given:Mean of 5 observations = 4Variance of 5 observations = 5.2Three observations: 1, 2, 6Let the other two observations be x and y.Mean: \(\frac{1 + 2 + 6 + x + y}{5} = 4\)⇒ \(9 + x + y = 20\)⇒ \(x + y = 11\) ...(1)Variance: \(\frac{1^2 + 2^2 + 6^2 + x^2 + y^2}{5} - 4^2 = 5.2\)⇒ \(\frac{1 + 4 + 36 + x^2 + y^2}{5} - 16 = 5.2\)⇒ \(\frac{41 + x^2 + y^2}{5} = 21.2\)⇒ \(41 + x^2 + y^2 = 106\)⇒ \(x^2 + y^2 = 65\) ...(2)From (1), \(y = 11 - x\)Substitute in (2):⇒ \(x^2 + (11 - x)^2 = 65\)⇒ \(x^2 + 121 - 22x + x^2 = 65\)⇒ \(2x^2 - 22x + 56 = 0\)⇒ \(x^2 - 11x + 28 = 0\)⇒ \((x - 4)(x - 7) = 0\)⇒ \(x = 4\) or \(x = 7\)If \(x = 4\), \(y = 11 - 4 = 7\)If \(x = 7\), \(y = 11 - 7 = 4\)The other two observations are 4 and 7.Hence option 1 is correct Statistics Question 2: For a statistical data x1, x2, …, x10 of 10 values, a student obtained the mean as 5.5 and \(\sum_{i=1}^{10} x_{i}^{2}=371\). He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is 7495Answer (Detailed Solution Below) Option 1 : 7 CalculationMean x̅ = 5.5 = \(\sum_{i=1}^{10} x_{i}=5.5 \times 10=55\)= \(\sum_{i=1}^{10} x_{i}^{2}=371\)\(\left(\sum \mathrm{x}_{\mathrm{i}}\right)_{\text {new }}=55-(4+5)+(6+8)=60\)\(\left(\sum \mathrm{x^2}_{\mathrm{i}}\right)_{\text {new }}=371-\left(4^{2}+5^{2}\right)+\left(6^{2}+8^{2}\right)=430\)Variance σ2 = \(\frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{10}-\left(\frac{\sum \mathrm{x}_{\mathrm{i}}}{10}\right)^{2}\)\(\sigma^{2}=\frac{430}{10}-\left(\frac{60}{10}\right)^{2}\)⇒ σ2 = 43 – 36⇒ σ2 = 7 Hence option 1 is correct Statistics Question 3: The variance of the numbers 8, 21, 34, 47, …, 320, is ________. Answer (Detailed Solution Below) 8788 Calculation8 + (n –1)13 = 320⇒ 13n = 325⇒ n = 25no. of terms = 25mean = \(\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}=\frac{8+21+\ldots+320}{25}=\frac{\frac{25}{2}(8+320)}{25}\)variance σ2 = \(\frac{\Sigma \mathrm{x}_{\mathrm{i}}^{2}}{\mathrm{n}}-(\text { mean })^{2}\)⇒ \(\frac{8^{2}+21^{2}+\ldots .+320^{2}}{13}-(164)^{2}\)⇒ 8788 Statistics Question 4: Marks obtains by all the students of class 12 are presented in a frequency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18, then the total number of students is