1. Masalani bayon qilish: Berilgan $f(x)$ uzluksiz tasodifiy miqdorning zichlik funksiyasi (PDF) sifatida berilgan. Bizga uning dastlabki to'rt markaziy momentlari va assimmetriya koeffitsienti $A_s$ ni topish vazifasi berilgan. 2. Formulalar va tushuntirishlar: - Dastlabki momentlar: $m_k = E[X^k] = \int_{-\infty}^{\infty} x^k f(x) dx$ - Markaziy momentlar: $\mu_k = E[(X - \mu)^k]$, bunda $\mu = m_1$ — birinchi moment (kutilgan qiymat). - Assimmetriya koeffitsienti: $A_s = \frac{\mu_3}{\mu_2^{3/2}}$ 3. $f(x)$ bo'yicha integral chegaralari va ifodalar: - $f(x) = 0$ agar $x < -4$ - $f(x) = \frac{2}{3} \left(1 + \frac{x}{4}\right)$ agar $-4 \leq x < 0$ - $f(x) = \frac{2}{3} \left(1 - \frac{x}{9}\right)$ agar $0 \leq x \leq 9$ - $f(x) = 0$ agar $x > 9$ 4. Birinchi momentni hisoblash: $$m_1 = \int_{-4}^0 x \cdot \frac{2}{3} \left(1 + \frac{x}{4}\right) dx + \int_0^9 x \cdot \frac{2}{3} \left(1 - \frac{x}{9}\right) dx$$ Hisoblaymiz birinchi integralni: $$\int_{-4}^0 x \cdot \frac{2}{3} \left(1 + \frac{x}{4}\right) dx = \frac{2}{3} \int_{-4}^0 \left(x + \frac{x^2}{4}\right) dx = \frac{2}{3} \left[ \frac{x^2}{2} + \frac{x^3}{12} \right]_{-4}^0 = \frac{2}{3} \left(0 - \left(\frac{16}{2} + \frac{-64}{12}\right)\right) = \frac{2}{3} \left(-8 + \frac{16}{3}\right) = \frac{2}{3} \left(-\frac{24}{3} + \frac{16}{3}\right) = \frac{2}{3} \left(-\frac{8}{3}\right) = -\frac{16}{9}$$ Ikkinchi integral: $$\int_0^9 x \cdot \frac{2}{3} \left(1 - \frac{x}{9}\right) dx = \frac{2}{3} \int_0^9 \left(x - \frac{x^2}{9}\right) dx = \frac{2}{3} \left[ \frac{x^2}{2} - \frac{x^3}{27} \right]_0^9 = \frac{2}{3} \left( \frac{81}{2} - \frac{729}{27} \right) = \frac{2}{3} \left(40.5 - 27\right) = \frac{2}{3} \times 13.5 = 9$$ Shunday qilib, $$m_1 = -\frac{16}{9} + 9 = -1.777... + 9 = 7.222... = \frac{65}{9}$$ 5. Ikkinchi momentni hisoblash: $$m_2 = \int_{-4}^0 x^2 \cdot \frac{2}{3} \left(1 + \frac{x}{4}\right) dx + \int_0^9 x^2 \cdot \frac{2}{3} \left(1 - \frac{x}{9}\right) dx$$ Birinchi integral: $$\frac{2}{3} \int_{-4}^0 \left(x^2 + \frac{x^3}{4}\right) dx = \frac{2}{3} \left[ \frac{x^3}{3} + \frac{x^4}{16} \right]_{-4}^0 = \frac{2}{3} \left(0 - \left(-\frac{64}{3} + \frac{256}{16}\right)\right) = \frac{2}{3} \left(\frac{64}{3} - 16\right) = \frac{2}{3} \left(\frac{64 - 48}{3}\right) = \frac{2}{3} \times \frac{16}{3} = \frac{32}{9}$$ Ikkinchi integral: $$\frac{2}{3} \int_0^9 \left(x^2 - \frac{x^3}{9}\right) dx = \frac{2}{3} \left[ \frac{x^3}{3} - \frac{x^4}{36} \right]_0^9 = \frac{2}{3} \left( \frac{729}{3} - \frac{6561}{36} \right) = \frac{2}{3} \left(243 - 182.25\right) = \frac{2}{3} \times 60.75 = 40.5$$ Shunday qilib, $$m_2 = \frac{32}{9} + 40.5 = 3.555... + 40.5 = 44.055... = \frac{397}{9}$$ 6. $\mu = m_1 = \frac{65}{9}$ ni hisobga olib, markaziy momentlarni topamiz: $$\mu_2 = m_2 - m_1^2 = \frac{397}{9} - \left(\frac{65}{9}\right)^2 = \frac{397}{9} - \frac{4225}{81} = \frac{3573}{81} - \frac{4225}{81} = -\frac{652}{81}$$ Bu manfiy chiqdi, demak hisobda xatolik bor. Qayta tekshirish kerak. 7. Qayta tekshirish uchun $m_1$ va $m_2$ ni aniqroq hisoblaymiz: $m_1$: $$-\frac{16}{9} + 9 = -1.777... + 9 = 7.222...$$ $m_2$: $$\frac{32}{9} + 40.5 = 3.555... + 40.5 = 44.055...$$ $\mu_2 = 44.055... - (7.222...)^2 = 44.055... - 52.17 = -8.11$$ Manfiy dispersiya mumkin emas, demak $m_1$ yoki $m_2$ hisobida xatolik bor. 8. To'g'ri hisoblash uchun integralni aniqroq bajarish: Birlamchi integral $m_1$ uchun: $$\int_{-4}^0 x \cdot \frac{2}{3} \left(1 + \frac{x}{4}\right) dx = \frac{2}{3} \int_{-4}^0 \left(x + \frac{x^2}{4}\right) dx = \frac{2}{3} \left[ \frac{x^2}{2} + \frac{x^3}{12} \right]_{-4}^0$$ Hisoblaymiz chegaralarda: $$\left(\frac{0^2}{2} + \frac{0^3}{12}\right) - \left(\frac{(-4)^2}{2} + \frac{(-4)^3}{12}\right) = 0 - \left(\frac{16}{2} - \frac{64}{12}\right) = 0 - (8 - 5.333...) = -2.666...$$ Shunday qilib, $$\frac{2}{3} \times (-2.666...) = -1.777... = -\frac{16}{9}$$ Ikkinchi integral $m_1$ uchun: $$\frac{2}{3} \int_0^9 \left(x - \frac{x^2}{9}\right) dx = \frac{2}{3} \left[ \frac{x^2}{2} - \frac{x^3}{27} \right]_0^9 = \frac{2}{3} \left(40.5 - 27\right) = 9$$ Shunday qilib, $m_1 = -\frac{16}{9} + 9 = \frac{65}{9} \approx 7.222$ to'g'ri. 9. $m_2$ uchun birinchi integral: $$\frac{2}{3} \int_{-4}^0 \left(x^2 + \frac{x^3}{4}\right) dx = \frac{2}{3} \left[ \frac{x^3}{3} + \frac{x^4}{16} \right]_{-4}^0 = \frac{2}{3} \left(0 - \left(-\frac{64}{3} + 16\right)\right) = \frac{2}{3} \left(\frac{64}{3} - 16\right) = \frac{2}{3} \times \frac{16}{3} = \frac{32}{9} \approx 3.555$$ Ikkinchi integral $m_2$ uchun: $$\frac{2}{3} \int_0^9 \left(x^2 - \frac{x^3}{9}\right) dx = \frac{2}{3} \left[ \frac{x^3}{3} - \frac{x^4}{36} \right]_0^9 = \frac{2}{3} \left(243 - 182.25\right) = 40.5$$ Shunday qilib, $$m_2 = \frac{32}{9} + 40.5 = 44.055...$$ 10. Endi markaziy dispersiyani hisoblaymiz: $$\mu_2 = m_2 - m_1^2 = 44.055... - (7.222...)^2 = 44.055... - 52.17 = -8.11$$ Bu manfiy chiqmoqda, demak $f(x)$ zichlik funksiyasi emas yoki hisobda xatolik bor. 11. Tekshirish uchun $f(x)$ ni tekshiramiz: $\int f(x) dx$ 1 ga teng bo'lishi kerak. $$\int_{-4}^0 \frac{2}{3} \left(1 + \frac{x}{4}\right) dx + \int_0^9 \frac{2}{3} \left(1 - \frac{x}{9}\right) dx$$ Birinchi integral: $$\frac{2}{3} \int_{-4}^0 \left(1 + \frac{x}{4}\right) dx = \frac{2}{3} \left[ x + \frac{x^2}{8} \right]_{-4}^0 = \frac{2}{3} (0 - (-4 + 2)) = \frac{2}{3} \times 2 = \frac{4}{3} > 1$$ Bu zichlik funksiyasi uchun noto'g'ri, chunki integral 1 dan katta. 12. Xulosa: Berilgan $f(x)$ zichlik funksiyasi emas, shuning uchun momentlar va assimmetriya hisoblash noto'g'ri. Agar $f(x)$ zichlik bo'lsa, yuqoridagi formulalar yordamida hisoblash mumkin edi.