Xo'sh, bu funktsiyaning ehtimollik zichligi funksiyasi ekanligini ko'rsatish uchun birinchi navbatda \(2 \le x \le 18\) diapazonida funktsiya musbat ekanligini va hamma joyda nolga teng bo'lishini va shuning uchun birinchi shart bajarilganligini ko'rishimiz mumkin.

Bu erda qilishimiz kerak bo'lgan asosiy narsa \(\int_{{ - \infty }}^{\infty }{{f\left( x \right)\,dx}} = 1\) ekanligini ko'rsatishdir.

\[\begin{align*}\int_{{ - \infty }}^{\infty }{{f\left( x \right)\,dx}} & = \int_{2}^{{18}}{{\frac{3}{{37760}}{x^2}\left( {20 - x} \right)\,dx}}\\ & = \frac{3}{{37760}}\int_{2}^{{18}}{{20{x^2} - {x^3}\,dx}} = \left. {\frac{3}{{37760}}\left( {\frac{{20}}{3}{x^3} - \frac{1}{4}{x^4}} \right)} \right|_2^{18} = 1\end{align*}\]

Integral bitta va shuning uchun bu aslida ehtimollik zichligi funktsiyasidir.

Avval shuni esda tutingki, funksiya nolga teng bo'lmagan \(x\) dagi chegaralarimiz tufayli bu \(P\left( {2 \le X \le 7} \right)\) ga teng. Mana bu muammo uchun ish.

\[P\left( {X \le 7} \right) = P\left( {2 \le X \le 7} \right) = \int_{2}^{7}{{\frac{3}{{37760}}{x^2}\left( {20 - x} \right)\,dx}} = \left. {\frac{3}{{37760}}\left( {\frac{{20}}{3}{x^3} - \frac{1}{4}{x^4}} \right)} \right|_2^7 = \require{bbox} \bbox[2pt,border:1px solid black]{{0.130065}}\]

E'tibor bering, biz birinchi qismda noaniq integralni allaqachon bajarganimizdan foydalanganmiz. Biz qilishimiz kerak bo'lgan yagona narsa, ushbu qismning chegaralariga mos keladigan chegaralarni o'zgartirish edi.

Avval shuni esda tutingki, funktsiya nolga teng bo'lmagan \(x\) dagi limitlarimiz tufayli bu \(P\left( {7 \le X \le 18} \right)\) ga teng. Mana bu muammo uchun ish.

\[P\left( {X \ge 7} \right) = P\left( {7 \le X \le 18} \right) = \int_{7}^{{18}}{{\frac{3}{{37760}}{x^2}\left( {20 - x} \right)\,dx}} = \left. {\frac{3}{{37760}}\left( {\frac{{20}}{3}{x^3} - \frac{1}{4}{x^4}} \right)} \right|_7^{18} = \require{bbox} \bbox[2pt,border:1px solid black]{{0.869935}}\]

E'tibor bering, biz birinchi qismda noaniq integralni allaqachon bajarganimizdan foydalanganmiz. Biz qilishimiz kerak bo'lgan yagona narsa, ushbu qismning chegaralariga mos keladigan chegaralarni o'zgartirish edi.

Bu erda integralni hisoblashdan boshqa ko'p narsa qilish kerak emas.

\[P\left( {3 \le X \le 14} \right) = \int_{3}^{{14}}{{\frac{3}{{37760}}{x^2}\left( {20 - x} \right)\,dx}} = \left. {\frac{3}{{37760}}\left( {\frac{{20}}{3}{x^3} - \frac{1}{4}{x^4}} \right)} \right|_3^{14} = \require{bbox} \bbox[2pt,border:1px solid black]{{0.677668}}\]

E'tibor bering, biz birinchi qismda noaniq integralni allaqachon bajarganimizdan foydalanganmiz. Biz qilishimiz kerak bo'lgan yagona narsa, ushbu qismning chegaralariga mos keladigan chegaralarni o'zgartirish edi.

Bu qism uchun faqat quyidagi integralni hisoblashimiz kerak.

\[\begin{align*}\mu & = \int_{{ - \infty }}^{\infty }{{x\,f\left( x \right)\,dx}} = \int_{2}^{{18}}{{x\left[ {\frac{3}{{37760}}{x^2}\left( {20 - x} \right)} \right]\,dx}}\\ & = \frac{3}{{37760}}\int_{2}^{{18}}{{20{x^3} - {x^4}\,dx}} = \left. {\frac{3}{{37760}}\left( {5{x^4} - \frac{1}{5}{x^5}} \right)} \right|_2^{18} = 11.6705\end{align*}\]

\(X\) ning o'rtacha qiymati keyin 11,6705 ni tashkil qiladi.