Statistik tavsiflar: o'rtacha, moda, mediana

Bu nima va nega kerak

Tasavvur qiling, futbol jamoasi so'nggi 5 ta o'yinda shuncha gol urdi: $1, 0, 5, 2, 2$. Jamoaning umumiy natijasini bitta son bilan qanday ifodalash mumkin?

Yoki ikki xil korxonadagi maoshlarni solishtirmoqchisiz: birida hamma 4 million so'm oladi, ikkinchisida 4 nafar xodim 2 million, rahbar esa 22 million so'm oladi. Qaysi ko'rsatkich oddiy xodim daromadini aniqroq aks ettiradi?

Mana shunday savollarga javob berish uchun statistika fani kerak. Statistika — sonli ma'lumotlarni yig'ish, tartibga solish va ulardan to'g'ri xulosa chiqarish vositasidir. Katta hajmdagi sonlarni yaxlit ifodalash uchun to'rtta asosiy tavsif ishlatiladi:

1. Arifmetik o'rtacha
2. Moda
3. Mediana
4. O'zgarish oralig'i (quloch)

Asosiy tushunchalar

1. Arifmetik o'rtacha

Ta'rif. Bir nechta sonning arifmetik o'rtachasi — bu barcha sonlar yig'indisini ularning soniga bo'lishdan hosil bo'lgan qiymatdir:

$\bar{x} = \frac{x_1 + x_2 + \dots + x_n}{n}$

Choraklik bahoni chiqarishda barcha baholar yig'ilib, baholar soniga bo'linadi.

2. Moda

Ta'rif. Ma'lumotlar to'plamida eng ko'p uchraydigan (takrorlanadigan) son moda ($M_o$) deyiladi.

• Agar bitta son ko'proq takrorlansa — to'plam yagona modaga ega: $2, 3, 3, 5 \implies M_o = 3$.
• Agar ikkita son bir xil eng ko'p takrorlansa — to'plam bimodal (ikki modali) bo'ladi: $2, 2, 4, 5, 5 \implies M_o = 2 \text{ va } 5$.
• Agar barcha sonlar bir xil miqdorda qatnashsa — to'plamning modasi yo'q.

Do'kon egasi uchun eng ko'p sotilgan poyabzal o'lchami modadir — u qaysi o'lchamdagi mahsulot eng xaridorgir ekanini ko'rsatadi.

3. Mediana

Ta'rif. O'sish tartibida yozilgan sonlar qatorining qoq o'rtasida turgan son mediana ($M_e$) deyiladi.

Qoidasi:

1. Sonlar to'plamini albatta o'sish tartibida saralang.
2. Agar sonlar soni $n$ toq bo'lsa, o'rtadagi $\dfrac{n+1}{2}$-o'rindagi son mediana bo'ladi.
3. Agar sonlar soni $n$ juft bo'lsa, o'rtada ikkita son turadi. Mediana mana shu o'rtadagi ikkita sonning arifmetik o'rtachasiga teng bo'ladi.

4. O'zgarish oralig'i (quloch)

Ta'rif. To'plamdagi eng katta va eng kichik qiymatning ayirmasi o'zgarish oralig'i deyiladi:

$R = x_{\text{max}} - x_{\text{min}}$

Masalan, eng baland harorat $+25^\circ\text{C}$, eng past harorat $+15^\circ\text{C}$ bo'lsa, oraliq $25 - 15 = 10^\circ\text{C}$ bo'ladi.

---

Yechilgan misollar

1-misol. Toq sondagi to'plamni to'liq tahlil qilish

Berilgan sonlar: $14, 7, 2, 7, 15$. Ushbu to'plamning arifmetik o'rtachasi, modasi, medianasi va o'zgarish oralig'ini toping.

• Tartiblash: $2, 7, 7, 14, 15$ ($n = 5$, toq son).
• Arifmetik o'rtacha: $\bar{x} = \dfrac{2 + 7 + 7 + 14 + 15}{5} = \dfrac{45}{5} = 9$.
• Moda: $7$ soni eng ko'p (2 marta) takrorlangan: $M_o = 7$.
• Mediana: 5 ta sonning o'rtasidagi 3-son — $7$. Demak, $M_e = 7$.
• O'zgarish oralig'i: $R = 15 - 2 = 13$.

Javob: O'rtacha $9$, moda $7$, mediana $7$, oraliq $13$.

2-misol. Juft sondagi to'plam

Olti kunlik savdo tushumlari (million so'mda): $12, 18, 16, 24, 30, 20$. Arifmetik o'rtacha va medianani toping.

• Tartiblash: $12, 16, 18, 20, 24, 30$ ($n = 6$, juft son).
• Arifmetik o'rtacha: $\bar{x} = \dfrac{12 + 16 + 18 + 20 + 24 + 30}{6} = \dfrac{120}{6} = 20$.
• Mediana: O'rtadagi 3- va 4-sonlar: $18$ va $20$. $M_e = \frac{18 + 20}{2} = 19$

Javob: Arifmetik o'rtacha $20$, mediana $19$.

3-misol. Noma'lum hadni arifmetik o'rtacha orqali topish

O'quvchi 4 ta imtihonda $82, 86, 90, 94$ ball oldi. 5-imtihondan keyin o'rtacha ball $90$ bo'lishi uchun u necha ball olishi kerak?

• Beshinchi ballni $x$ deb belgilaymiz: $\frac{82 + 86 + 90 + 94 + x}{5} = 90$ $\frac{352 + x}{5} = 90 \implies 352 + x = 450 \implies x = 98$

Javob: O'quvchi $98$ ball olishi kerak.

4-misol. Chastotali jadval

10 nafar o'quvchining yechgan masalalari soni jadvalda berilgan:

Masalalar soni ($x$)	O'quvchilar soni ($f$)
$3$	$3$
$4$	$4$
$5$	$3$

• Arifmetik o'rtacha: $\bar{x} = \dfrac{3\cdot 3 + 4\cdot 4 + 3\cdot 5}{3 + 4 + 3} = \dfrac{9 + 16 + 15}{10} = \dfrac{40}{10} = 4$.
• Moda: Eng katta chastota $4$ ta o'quvchi, unga mos qiymat $4$. Demak, $M_o = 4$.
• Mediana: Qator: $3, 3, 3, 4, 4, 4, 4, 5, 5, 5$. O'rtadagi 5- va 6-sonlar $4$ ga teng. $M_e = 4$.

Javob: O'rtacha $4$, moda $4$, mediana $4$.

5-misol. Anomaliya va mediananing afzalligi

Besh nafar xodimning oylik maoshi: $3, 3, 4, 5, 45$ (million so'mda).

• Arifmetik o'rtacha: $\bar{x} = \dfrac{3 + 3 + 4 + 5 + 45}{5} = \dfrac{60}{5} = 12$ mln so'm.
• Mediana: Qator o'rtasidagi son: $M_e = 4$ mln so'm.

Arifmetik o'rtacha ($12$ mln) bitta keskin katta qiymat ($45$ mln) tufayli haqiqatni buzib ko'rsatadi. Mediana ($4$ mln) esa jamoaning haqiqiy holatini adolatliroq ifodalaydi.

---

Mashq

1. $5, 8, 11, 14, 17$ sonlarining arifmetik o'rtachasini toping.
2. $4, 6, 9, 4, 7, 4, 6, 8, 10$ to'plamning modasini toping.
3. $14, 9, 21, 6, 17, 11, 27$ sonlar qatorining medianasini toping.
4. $2, 6, 8, 12, 16, 20, 24, 28$ qatorining medianasini toping.
5. $15, 23, 8, 41, 19, 32, 11$ sonlar to'plamining o'zgarish oralig'ini toping.
6. O'quvchining 4 ta testdagi ballari: $78, 82, 86, 90$. 5-testda necha ball olsa, o'rtacha bali $85$ bo'ladi?
7. $2, 5, 5, 7, 9, 9, 10, 17$ qatorining arifmetik o'rtachasi, modasi va medianasini toping.
8. 10 kunlik savdo hajmi: $15, 20, 25, 30, 30, 30, 35, 40, 45, 50$. O'rtacha, moda, mediana va oraliqni toping.
9. $12, 15, 18, 21, 24, x$ sonlarining arifmetik o'rtachasi $20$ ga teng. $x$ ni toping.
10. O'sish tartibida yozilgan $3, 7, x, 16, 18$ qatorida mediana arifmetik o'rtachaga teng. $x$ ni toping.

Javoblar va yechim yo'llari

1. $\bar{x} = \dfrac{5+8+11+14+17}{5} = \dfrac{55}{5} = 11$. Javob: 11.
2. $4$ soni $3$ marta, $6$ soni $2$ marta, qolganlari $1$ marta. Eng ko'p takrorlangan son $4$. Javob: 4.
3. Tartiblangan qator: $6, 9, 11, \mathbf{14}, 17, 21, 27$. O'rtadagi to'rtinchi son $14$. Javob: 14.
4. O'rtadagi 4- va 5-sonlar $12$ va $16$. Mediana: $\dfrac{12 + 16}{2} = 14$. Javob: 14.
5. Eng katta $41$, eng kichik $8$. Oraliq: $41 - 8 = 33$. Javob: 33.
6. 4 ta test yig'indisi: $78 + 82 + 86 + 90 = 336$. Kerakli yig'indi: $5 \cdot 85 = 425$. Beshinchi test: $425 - 336 = 89$. Javob: 89.
7. Yig'indi $64$, o'rtacha: $\dfrac{64}{8} = 8$. Modalar: $5$ va $9$ (har biri 2 martadan). Mediana: $\dfrac{7+9}{2} = 8$. Javob: o'rtacha 8, modalar 5 va 9, mediana 8.
8. Yig'indi $320$, o'rtacha: $\dfrac{320}{10} = 32$. Moda: $30$. Mediana: $\dfrac{30+30}{2} = 30$. Oraliq: $50 - 15 = 35$. Javob: o'rtacha 32, moda 30, mediana 30, oraliq 35.
9. Barcha 6 ta son yig'indisi: $6 \cdot 20 = 120$. Ma'lum sonlar yig'indisi $90$. $x = 120 - 90 = 30$. Javob: 30.
10. Mediana $x$ ga teng. O'rtacha: $\dfrac{3 + 7 + x + 16 + 18}{5} = \dfrac{44 + x}{5}$. Shart bo'yicha: $x = \dfrac{44 + x}{5} \implies 5x = 44 + x \implies 4x = 44 \implies x = 11$. Javob: 11.

---

Keng tarqalgan xatolar

1. Medianani topishdan oldin qatorni saralamaslik. ✗ $7, 2, 9, 1, 5$ qatorida o'rtada $9$ turibdi, demak mediana $9$. ✓ Avval o'sish tartibida yozish shart: $1, 2, \mathbf{5}, 7, 9$. O'rtadagi son $5$, demak mediana $5$.

2. Juft sonli qatorda medianani shunchaki bitta songa teng deb olish. ✗ $2, 4, 6, 8$ qatorida mediana $4$ (yoki $6$). ✓ Juft sondagi qatorda o'rtadagi ikkita sonning arifmetik o'rtachasi olinadi: $\dfrac{4+6}{2} = 5$.

3. Moda yo'qligini $0$ ga teng deb hisoblash. ✗ $3, 7, 8, 12$ to'plamda hech qaysi son takrorlanmagan, demak moda $0$. ✓ Agar takrorlanish bo'lmasa, to'plamda "moda yo'q" deyiladi. $0$ esa alohida son qiymatidir.

4. Arifmetik o'rtachada bo'luvchini adashtirish. ✗ $4, 6, 8$ sonlari arifmetik o'rtachasi: $(4+6+8) : 2 = 9$. ✓ Hadlar soni $3$ ta bo'lgani uchun $3$ ga bo'linadi: $(4+6+8) : 3 = 6$.

---

DTM uslubidagi test

Ushbu mavzuni mustahkamlash uchun berilgan 5 ta test savolini ishlab, javoblaringizni tekshiring.

Keyingi qadam

Statistik tavsiflar bilan tanishdik. Keyingi bosqichlarda ushbu ko'rsatkichlar asosida ma'lumotlar diagrammalari va gistogrammalar tuzishni o'rganamiz.