8.5-bo'lim : Ehtimollik
2. Lampochka markasi uchun lampochkaning ishlash muddatining ehtimollik zichligi funksiyasi quyidagi funktsiya bilan berilgan, bu erda t oylarda.
\[f\left( t \right) = \left\{ {\begin{array}{ll}{0.04{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}}&{{\mbox{if }}t \ge 0}\\0&{{\mbox{if }}t < 0}\end{array}} \right.\]- \(f\left( t \right)\) ehtimollik zichligi funksiyasi ekanligini tekshiring.
- Lampochkaning ishlash muddati 8 oydan kam bo'lishi ehtimoli qanday?
- Lampochkaning 20 oydan ortiq xizmat qilish ehtimoli qanday?
- Lampochkaning 14 oydan 30 oygacha xizmat qilish ehtimoli qanday?
- Lampochkalarning ishlash muddatining o'rtacha qiymatini aniqlang.
a \(f\left( t \right)\) ehtimollik zichligi funksiyasi ekanligini tekshiring. Yechim
Xo‘sh, bu funksiya ehtimollik zichligi funksiyasi ekanligini ko‘rsatish uchun birinchi navbatda biz ulagan \(t\) qiymatidan qat’iy nazar eksponensial qism har doim ijobiy bo‘lishini va funksiyaning qolgan qismi har doim nolga teng ekanligini 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.
\[\begin{align*}\int_{{ - \infty }}^{\infty }{{f\left( x \right)\,dx}} & = \int_{0}^{\infty }{{0.04{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}} = \mathop {\lim }\limits_{n \to \infty } \int_{0}^{n}{{0.04{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}}\\ & = \mathop {\lim }\limits_{n \to \infty } \left. {\left( { - {{\bf{e}}^{ - \,\,\frac{t}{{25}}}}} \right)} \right|_0^n = \mathop {\lim }\limits_{n \to \infty } \left( { - {{\bf{e}}^{ - \,\,\frac{n}{{25}}}} + 1} \right) = 0 + 1 = 1\end{align*}\]Integral bitta va shuning uchun bu aslida ehtimollik zichligi funktsiyasidir.
Ushbu integral uchun cheksiz chegara bilan to'g'ri munosabatda bo'lishni unutmang! Agar bu turdagi integrallar bilan qanday ishlashni eslamasangiz, ga qayting Noto'g'ri integral bo'lim va tezda ko'rib chiqing!
b Lampochkaning ishlash muddati 8 oydan kam bo'lishi ehtimoli qanday? Yechim
Bu muammo bizdan hisoblashni so'ragan narsa - \(P\left( {X \le 8} \right)\). Bundan tashqari, funktsiya nolga teng bo'lmagan \(t\) chegaralarimiz tufayli bu \(P\left( {0 \le X \le 8} \right)\) ga teng. Mana bu muammo uchun ish.
\[P\left( {X \le 8} \right) = P\left( {0 \le X \le 8} \right) = \int_{0}^{8}{{0.04{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}} = \left. { - {{\bf{e}}^{ - \,\,\frac{t}{{25}}}}} \right|_0^8 = \require{bbox} \bbox[2pt,border:1px solid black]{{0.273851}}\]c Lampochkaning 20 oydan ortiq xizmat qilish ehtimoli qanday? Yechim
Bu muammo bizdan hisoblashni so'ragan narsa \(P\left( {X \ge 20} \right)\). Mana bu muammo uchun ish.
\[\begin{align*}P\left( {X \ge 20} \right) & = \int_{{20}}^{\infty }{{0.04{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}} = \mathop {\lim }\limits_{n \to \infty } \int_{{20}}^{n}{{0.04{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}}\\ & = \mathop {\lim }\limits_{n \to \infty } \left. {\left( { - {{\bf{e}}^{ - \,\,\frac{t}{{25}}}}} \right)} \right|_{20}^n = \mathop {\lim }\limits_{n \to \infty } \left( { - {{\bf{e}}^{ - \,\,\frac{n}{{25}}}} + {{\bf{e}}^{ - \,\,\frac{{20}}{{25}}}}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{0.449329}}\end{align*}\]Ushbu integral uchun cheksiz chegara bilan to'g'ri munosabatda bo'lishni unutmang! Agar bu turdagi integrallar bilan qanday ishlashni eslamasangiz, ga qayting Noto'g'ri integral bo'lim va tezda ko'rib chiqing!
d Lampochkaning 14 oydan 30 oygacha xizmat qilish ehtimoli qanday? Yechim
Bu muammo bizdan hisoblashni so'ragan narsa \(P\left( {14 \le X \le 30} \right)\). Mana bu muammo uchun ish.
\[P\left( {14 \le X \le 30} \right) = \int_{{14}}^{{30}}{{0.04{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}} = \left. { - {{\bf{e}}^{ - \,\,\frac{t}{{25}}}}} \right|_{14}^{30} = \require{bbox} \bbox[2pt,border:1px solid black]{{0.270015}}\]e Lampochkalarning ishlash muddatining o'rtacha qiymatini aniqlang. Yechim
Bu qism uchun faqat quyidagi integralni hisoblashimiz kerak.
\[\begin{align*}\mu & = \int_{{ - \infty }}^{\infty }{{t\,f\left( t \right)\,dt}} = \int_{0}^{\infty }{{0.04t{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}} = \mathop {\lim }\limits_{n \to \infty } \int_{0}^{n}{{0.04t{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}}\\ & = \mathop {\lim }\limits_{n \to \infty } \left. {\left[ { - t{{\bf{e}}^{ - \,\,\frac{t}{{25}}}} - 25{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}} \right]} \right|_0^n = \mathop {\lim }\limits_{n \to \infty } \left[ { - n{{\bf{e}}^{ - \,\,\frac{n}{{25}}}} - 25{{\bf{e}}^{ - \,\,\frac{n}{{25}}}} - \left( { - 25} \right)} \right]\\ & = \mathop {\lim }\limits_{n \to \infty } \left[ { - \frac{n}{{{{\bf{e}}^{\frac{n}{{25}}}}}} - 25{{\bf{e}}^{ - \,\,\frac{n}{{25}}}} + 25} \right] = \mathop {\lim }\limits_{n \to \infty } \left[ { - \frac{1}{{\frac{1}{{25}}{{\bf{e}}^{\frac{n}{{25}}}}}}} \right] - 25\left( 0 \right) + 25 = \require{bbox} \bbox[2pt,border:1px solid black]{{25}}\end{align*}\]Lampochkalarning ishlash muddatining o'rtacha qiymati keyin 25 oyni tashkil qiladi.
Integralni bajarish uchun qismlar bo'yicha integratsiyadan foydalanishimiz kerak edi. Agar ko'rish kerak bo'lsa, bu ish.
\[\begin{align*} & \int{{0.04t{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}u = 0.04t\,\,\,\,\,\,du = 0.04dt\,\,\,\,\,\,\,\,\,\,\,\,dv = {{\bf{e}}^{ - \,\,\frac{t}{{25}}}}dt\,\,\,\,\,\,v = - 25{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\\ & \int{{0.04t{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}} = - t{{\bf{e}}^{ - \,\,\frac{t}{{25}}}} + \int{{{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}} = - t{{\bf{e}}^{ - \,\,\frac{t}{{25}}}} - 25{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\end{align*}\]Bundan tashqari, birinchi muddat chegarasi uchun biz limitni bajarish uchun L'Hospital qoidasidan foydalandik.