{"id":368049,"date":"2024-07-21T15:17:15","date_gmt":"2024-07-21T08:17:15","guid":{"rendered":"https:\/\/hoanghamobile.com\/tin-tuc\/?p=368049"},"modified":"2024-07-21T15:17:15","modified_gmt":"2024-07-21T08:17:15","slug":"trung-binh-cong-la-gi","status":"publish","type":"post","link":"https:\/\/hoanghamobile.com\/tin-tuc\/trung-binh-cong-la-gi\/","title":{"rendered":"Trung b\u00ecnh c\u1ed9ng l\u00e0 g\u00ec? C\u00f4ng th\u1ee9c t\u00ednh trung b\u00ecnh c\u1ed9ng chu\u1ea9n nh\u1ea5t?"},"content":{"rendered":"

Trung b\u00ecnh c\u1ed9ng l\u00e0 m\u1ed9t ph\u00e9p to\u00e1n c\u01a1 b\u1ea3n trong to\u00e1n h\u1ecdc, gi\u00fap ta t\u00ednh to\u00e1n gi\u00e1 tr\u1ecb trung b\u00ecnh c\u1ee7a c\u00e1c t\u1eadp h\u1ee3p s\u1ed1. B\u00ean c\u1ea1nh \u0111\u00f3, n\u00f3 c\u0169ng \u0111\u01b0\u1ee3c \u1ee9ng d\u1ee5ng r\u1ed9ng r\u00e3i trong nhi\u1ec1u l\u0129nh v\u1ef1c, t\u1eeb khoa h\u1ecdc, k\u1ef9 thu\u1eadt \u0111\u1ebfn kinh t\u1ebf, x\u00e3 h\u1ed9i v\u00e0 \u0111\u1eddi s\u1ed1ng h\u00e0ng ng\u00e0y. V\u1eady trung b\u00ecnh c\u1ed9ng l\u00e0 g\u00ec<\/strong>? C\u00f4ng th\u1ee9c t\u00ednh trung b\u00ecnh c\u1ed9ng ra sao? B\u00e0i vi\u1ebft n\u00e0y s\u1ebd cung c\u1ea5p cho b\u1ea1n nh\u1eefng ki\u1ebfn th\u1ee9c c\u01a1 b\u1ea3n v\u1ec1 trung b\u00ecnh c\u1ed9ng bao g\u1ed3m \u0111\u1ecbnh ngh\u0129a, c\u00f4ng th\u1ee9c t\u00ednh v\u00e0 m\u1ed9t s\u1ed1 v\u00ed d\u1ee5 minh h\u1ecda. \u0110\u1ed3ng th\u1eddi, b\u1ea1n c\u0169ng s\u1ebd \u0111\u01b0\u1ee3c h\u01b0\u1edbng d\u1eabn c\u00e1ch \u00e1p d\u1ee5ng trung b\u00ecnh c\u1ed9ng v\u00e0o gi\u1ea3i quy\u1ebft c\u00e1c b\u00e0i to\u00e1n th\u1ef1c t\u1ebf trong nhi\u1ec1u l\u0129nh v\u1ef1c kh\u00e1c nhau.<\/p>\n

Trung b\u00ecnh c\u1ed9ng l\u00e0 g\u00ec?<\/h2>\n

Trung b\u00ecnh c\u1ed9ng (hay c\u00f2n g\u1ecdi l\u00e0 trung b\u00ecnh s\u1ed1 h\u1ecdc) l\u00e0 th\u01b0\u01a1ng s\u1ed1 gi\u1eefa t\u1ed5ng gi\u00e1 tr\u1ecb c\u1ee7a m\u1ed9t t\u1eadp h\u1ee3p s\u1ed1 v\u00e0 s\u1ed1 l\u01b0\u1ee3ng c\u00e1c s\u1ed1 trong t\u1eadp h\u1ee3p \u0111\u00f3. N\u00f3i c\u00e1ch kh\u00e1c, trung b\u00ecnh c\u1ed9ng l\u00e0 gi\u00e1 tr\u1ecb \u0111\u1ea1i di\u1ec7n cho m\u1ee9c \u0111\u1ed9 t\u1eadp trung c\u1ee7a c\u00e1c s\u1ed1 trong t\u1eadp h\u1ee3p.<\/p>\n

C\u00f4ng th\u1ee9c t\u00ednh trung b\u00ecnh c\u1ed9ng:<\/strong> Trung b\u00ecnh c\u1ed9ng = T\u1ed5ng gi\u00e1 tr\u1ecb \/ S\u1ed1 l\u01b0\u1ee3ng s\u1ed1<\/strong><\/em><\/p>\n

V\u00ed d\u1ee5: <\/strong>Gi\u1ea3 s\u1eed ta c\u00f3 t\u1eadp h\u1ee3p s\u1ed1 A = {1, 2, 3, 4, 5}. Trung b\u00ecnh c\u1ed9ng c\u1ee7a t\u1eadp h\u1ee3p A \u0111\u01b0\u1ee3c t\u00ednh nh\u01b0 sau:<\/p>\n

Trung b\u00ecnh c\u1ed9ng (A) = (1 + 2 + 3 + 4 + 5) \/ 5 = 15 \/ 5 = 3<\/p>\n

\"cong-thuc-tinh-trung-binh-cong-la-gi-6\"<\/picture><\/p>\n

L\u01b0u \u00fd r\u1eb1ng trung b\u00ecnh c\u1ed9ng ch\u1ec9 n\u00ean \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng cho c\u00e1c t\u1eadp h\u1ee3p s\u1ed1 c\u00f3 gi\u00e1 tr\u1ecb g\u1ea7n nhau. N\u1ebfu t\u1eadp h\u1ee3p s\u1ed1 c\u00f3 c\u00e1c gi\u00e1 tr\u1ecb kh\u00e1c bi\u1ec7t l\u1edbn, vi\u1ec7c s\u1eed d\u1ee5ng trung b\u00ecnh c\u1ed9ng c\u00f3 th\u1ec3 kh\u00f4ng ph\u1ea3n \u00e1nh ch\u00ednh x\u00e1c m\u1ee9c \u0111\u1ed9 t\u1eadp trung c\u1ee7a c\u00e1c s\u1ed1 trong t\u1eadp h\u1ee3p. Ngo\u00e0i trung b\u00ecnh c\u1ed9ng, c\u00f2n c\u00f3 c\u00e1c lo\u1ea1i trung b\u00ecnh kh\u00e1c nh\u01b0 trung b\u00ecnh nh\u00e2n, trung b\u00ecnh tuy\u1ebfn t\u00ednh,… \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng trong c\u00e1c tr\u01b0\u1eddng h\u1ee3p c\u1ee5 th\u1ec3.<\/p>\n

M\u1ed9t v\u00e0i b\u00e0i to\u00e1n t\u00ecm trung b\u00ecnh c\u1ed9ng<\/h2>\n

Trong n\u1ed9i dung ti\u1ebfp theo ch\u00fang ta s\u1ebd t\u1eadp trung gi\u1ea3i m\u1ed9t s\u1ed1 b\u00e0i to\u00e1n t\u00ecm trung b\u00ecnh c\u1ed9ng th\u01b0\u1eddng g\u1eb7p trong ch\u01b0\u01a1ng tr\u00ecnh h\u1ecdc To\u00e1n l\u1edbp 4. C\u00e1c b\u00e0i to\u00e1n n\u00e0y \u0111\u01b0\u1ee3c tr\u00ecnh b\u00e0y theo m\u1ee9c \u0111\u1ed9 kh\u00f3 d\u1ec5 t\u0103ng d\u1ea7n, gi\u00fap h\u1ecdc sinh d\u1ec5 d\u00e0ng ti\u1ebfp c\u1eadn v\u00e0 r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng t\u00ednh to\u00e1n trung b\u00ecnh c\u1ed9ng. B\u1eaft \u0111\u1ea7u v\u1edbi nh\u1eefng b\u00e0i to\u00e1n \u0111\u01a1n gi\u1ea3n nh\u1ea5t, ta s\u1ebd t\u00ecm hi\u1ec3u c\u00e1ch t\u00ednh trung b\u00ecnh c\u1ed9ng c\u1ee7a hai s\u1ed1 v\u00e0 nhi\u1ec1u s\u1ed1. Sau \u0111\u00f3, ta s\u1ebd chuy\u1ec3n sang nh\u1eefng b\u00e0i to\u00e1n ph\u1ee9c t\u1ea1p h\u01a1n, \u00e1p d\u1ee5ng t\u00ednh trung b\u00ecnh c\u1ed9ng v\u00e0o c\u00e1c t\u00ecnh hu\u1ed1ng th\u1ef1c t\u1ebf trong \u0111\u1eddi s\u1ed1ng.<\/p>\n

Nh\u1eefng b\u00e0i to\u00e1n c\u01a1 b\u1ea3n sau s\u1ebd ch\u1ec9 t\u1eadp trung \u00e1p d\u1ee5ng c\u00f4ng th\u1ee9c m\u00e0 ch\u00fang ta \u0111\u00e3 \u0111\u01b0\u1ee3c t\u00ecm hi\u1ec3u trong n\u1ed9i dung trung b\u00ecnh c\u1ed9ng l\u00e0 g\u00ec \u1edf tr\u00ean.<\/p>\n

B\u00e0i to\u00e1n 1<\/h3>\n

a)\u00a0<\/strong>Lan c\u00f3 15 vi\u00ean k\u1eb9o, Hoa c\u00f3 18 vi\u00ean k\u1eb9o. Trung b\u00ecnh m\u1ed7i b\u1ea1n c\u00f3 bao nhi\u00eau vi\u00ean k\u1eb9o?<\/p>\n

Gi\u1ea3i: <\/strong>Trung b\u00ecnh c\u1ed9ng s\u1ed1 k\u1eb9o c\u1ee7a hai b\u1ea1n l\u00e0: (15 + 18) \/ 2 = 16,5 (vi\u00ean k\u1eb9o)<\/p>\n

V\u1eady trung b\u00ecnh m\u1ed7i b\u1ea1n c\u00f3 16,5 vi\u00ean k\u1eb9o.<\/p>\n

b)\u00a0<\/strong>Tr\u00ean b\u00e0n c\u00f3 5 qu\u1ea3 t\u00e1o v\u00e0 7 qu\u1ea3 cam. Trung b\u00ecnh m\u1ed7i lo\u1ea1i tr\u00e1i c\u00e2y c\u00f3 bao nhi\u00eau qu\u1ea3?<\/p>\n

Gi\u1ea3i: <\/strong>T\u1ed5ng s\u1ed1 tr\u00e1i c\u00e2y tr\u00ean b\u00e0n l\u00e0: (5 + 7) \/ 2 = 6 (qu\u1ea3)<\/p>\n

V\u1eady trung b\u00ecnh m\u1ed7i lo\u1ea1i tr\u00e1i c\u00e2y c\u00f3 6 qu\u1ea3.<\/p>\n

B\u00e0i to\u00e1n 2<\/h3>\n

\u0110\u1ec1 b\u00e0i:<\/strong> Ta bi\u1ebft \u0111\u01b0\u1ee3c t\u1ed5ng c\u00e1c s\u1ed1 h\u1ea1ng l\u00e0 180. Trung b\u00ecnh c\u1ed9ng c\u1ee7a ch\u00fang l\u00e0 60. H\u1ecfi trung b\u00ecnh c\u1ed9ng \u0111\u00f3 l\u00e0 c\u1ee7a bao nhi\u00eau s\u1ed1 h\u1ea1ng?<\/p>\n

Gi\u1ea3i:<\/strong> T\u1ed5ng \u0111\u00f3 c\u00f3 s\u1ed1 s\u1ed1 h\u1ea1ng l\u00e0: 180 : 60 = 3<\/p>\n

\"cong-thuc-tinh-trung-binh-cong-la-gi-3\"<\/picture><\/p>\n

K\u1ec3 t\u1eeb b\u00e0i to\u00e1n 3 tr\u1edf \u0111i ch\u00fang ta v\u1eabn s\u1ebd d\u00f9ng c\u00f4ng th\u1ee9c \u0111\u00e3 \u0111\u1ec1 c\u1eadp trong n\u1ed9i dung t\u00ecm hi\u1ec3u trung b\u00ecnh c\u1ed9ng l\u00e0 g\u00ec nh\u01b0ng s\u1ebd “bi\u1ebfn h\u00f3a” m\u1ed9t ch\u00fat.<\/p>\n

B\u00e0i to\u00e1n 3<\/h3>\n

\u0110\u1ec1 b\u00e0i:<\/strong> L\u1edbp 6A c\u00f3 40 h\u1ecdc sinh. \u0110i\u1ec3m ki\u1ec3m tra m\u00f4n To\u00e1n c\u1ee7a c\u00e1c b\u1ea1n h\u1ecdc sinh n\u00e0y \u0111\u01b0\u1ee3c th\u1ed1ng k\u00ea nh\u01b0 sau:<\/p>\n

\n
\n
\n
\n\n\n\n\n\n\n\n\n
\u0110i\u1ec3m<\/th>\nS\u1ed1 h\u1ecdc sinh<\/th>\n<\/tr>\n
7<\/td>\n8<\/td>\n<\/tr>\n
8<\/td>\n12<\/td>\n<\/tr>\n
9<\/td>\n10<\/td>\n<\/tr>\n
10<\/td>\n6<\/td>\n<\/tr>\n
11<\/td>\n4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

H\u00e3y t\u00ednh \u0111i\u1ec3m trung b\u00ecnh m\u00f4n To\u00e1n c\u1ee7a h\u1ecdc sinh l\u1edbp 6A.<\/p>\n

Gi\u1ea3i:<\/strong><\/p>\n

B\u01b0\u1edbc 1:<\/strong> T\u00ednh t\u1ed5ng \u0111i\u1ec3m c\u1ee7a c\u00e1c b\u1ea1n h\u1ecdc sinh: 7 x 8 + 8 x 12 + 9 x 10 + 10 x 6 + 11 x 4 = 236 (\u0111i\u1ec3m)<\/p>\n

B\u01b0\u1edbc 2:<\/strong> T\u00ednh s\u1ed1 h\u1ecdc sinh: 8 + 12 + 10 + 6 + 4 = 40 (h\u1ecdc sinh)<\/p>\n

B\u01b0\u1edbc 3:<\/strong> T\u00ednh \u0111i\u1ec3m trung b\u00ecnh: 236 \/ 40 = 5,9 (\u0111i\u1ec3m)<\/p>\n

V\u1eady \u0111i\u1ec3m trung b\u00ecnh m\u00f4n To\u00e1n c\u1ee7a h\u1ecdc sinh l\u1edbp 6A l\u00e0 5,9.<\/p>\n

B\u00e0i to\u00e1n 4<\/h3>\n

\u0110\u1ec1 b\u00e0i:<\/strong> Hai nh\u00f3m h\u1ecdc sinh A v\u00e0 B tham gia thi h\u1ecdc sinh gi\u1ecfi. Nh\u00f3m A c\u00f3 15 h\u1ecdc sinh, \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a nh\u00f3m l\u00e0 8,2. Nh\u00f3m B c\u00f3 18 h\u1ecdc sinh, \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a nh\u00f3m l\u00e0 7,8. H\u00e3y t\u00ednh \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a t\u1ea5t c\u1ea3 h\u1ecdc sinh tham gia thi.<\/p>\n

Gi\u1ea3i:<\/strong><\/p>\n

G\u1ecdi \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a t\u1ea5t c\u1ea3 h\u1ecdc sinh tham gia thi l\u00e0 M<\/span><\/span><\/span><\/span>.<\/p>\n

T\u1ed5ng s\u1ed1 \u0111i\u1ec3m c\u1ee7a nh\u00f3m A l\u00e0: 15 x <\/span><\/span>8<\/span>,<\/span>2 <\/span>= <\/span><\/span>123<\/span><\/span><\/span><\/span><\/span><\/p>\n

T\u1ed5ng s\u1ed1 \u0111i\u1ec3m c\u1ee7a nh\u00f3m B l\u00e0: 18 x <\/span><\/span>7<\/span>,<\/span>8 <\/span>= <\/span><\/span>140<\/span>,<\/span>4<\/span><\/span><\/span><\/span><\/span><\/p>\n

T\u1ed5ng s\u1ed1 \u0111i\u1ec3m c\u1ee7a t\u1ea5t c\u1ea3 h\u1ecdc sinh tham gia thi l\u00e0: 123 <\/span>+ <\/span><\/span>140<\/span>,<\/span>4 <\/span>= <\/span><\/span>263<\/span>,<\/span>4<\/span><\/span><\/span><\/span><\/span><\/p>\n

T\u1ed5ng s\u1ed1 h\u1ecdc sinh tham gia thi l\u00e0: 15 <\/span>+ <\/span><\/span>18 <\/span>= <\/span><\/span>33<\/span><\/span><\/span><\/span><\/span><\/p>\n

V\u1eady \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a t\u1ea5t c\u1ea3 h\u1ecdc sinh tham gia thi l\u00e0: M<\/span>=<\/span><\/span>263,<\/span>4<\/span><\/span>\u200b \/ 33 <\/span><\/span><\/span><\/span><\/span>= <\/span><\/span>7<\/span>,<\/span>98<\/span><\/span><\/span><\/span><\/span><\/p>\n

V\u1eady \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a t\u1ea5t c\u1ea3 h\u1ecdc sinh tham gia thi l\u00e0 7,98.<\/p>\n

\"cong-thuc-tinh-trung-binh-cong-la-gi-1\"<\/picture><\/p>\n

Sau \u0111\u00e2y l\u00e0 m\u1ed9t s\u1ed1 b\u00e0i to\u00e1n th\u1ef1c t\u1ebf \u0111\u1ec3 ch\u00fang ta c\u00f3 th\u1ec3 luy\u1ec7n t\u1eadp v\u00e0 hi\u1ec3u r\u00f5 h\u01a1n c\u00f4ng th\u1ee9c t\u00ednh trung b\u00ecnh c\u1ed9ng l\u00e0 g\u00ec.<\/p>\n

B\u00e0i to\u00e1n 5<\/h3>\n

\u0110\u1ec1 b\u00e0i: <\/strong>An c\u00f3 36 b\u00f4ng hoa. Nhi c\u00f3 nhi\u1ec1u h\u01a1n An 2 b\u00f4ng hoa nh\u01b0ng l\u1ea1i \u00edt h\u01a1n Nhung 2 b\u00f4ng hoa. H\u1ecfi trung b\u00ecnh b\u1ea1n c\u00f3 bao nhi\u00eau b\u00f4ng hoa?<\/p>\n

Gi\u1ea3i:<\/strong><\/p>\n

S\u1ed1 b\u00f4ng hoa m\u00e0 Nhi c\u00f3 l\u00e0:\u00a036 + 2 = 38 (b\u00f4ng hoa)<\/p>\n

S\u1ed1 b\u00f4ng hoa m\u00e0 Nhung c\u00f3 l\u00e0:\u00a038 + 2 = 40 (b\u00f4ng hoa)<\/p>\n

Trung b\u00ecnh m\u1ed7i b\u1ea1n c\u00f3 s\u1ed1 b\u00f4ng hoa l\u00e0:\u00a0(36 + 38 + 40) \/ 3 = 38 (b\u00f4ng hoa)<\/p>\n

V\u1eady trung b\u00ecnh m\u1ed7i b\u1ea1n c\u00f3 38 b\u00f4ng hoa.<\/p>\n

B\u00e0i to\u00e1n 6<\/h3>\n

\u0110\u1ec1 b\u00e0i:\u00a0<\/strong>M\u1ed9t c\u1eeda h\u00e0ng c\u00e2y c\u1ea3nh di chuy\u1ec3n c\u00e1c ch\u1eadu hoa h\u1ed3ng b\u1eb1ng xe \u0111\u1ea9y. L\u1ea7n \u0111\u1ea7u c\u00f3 3 xe \u0111\u1ea9y, m\u1ed7i xe \u0111\u1ea9y chuy\u1ec3n \u0111\u01b0\u1ee3c 16 ch\u1eadu. Sau \u0111\u00f3 c\u00f3 5 xe \u0111\u1ea9y kh\u00e1c l\u1edbn h\u01a1n v\u00e0 m\u1ed7i xe \u0111\u1ea9y chuy\u1ec3n \u0111\u01b0\u1ee3c 24 ch\u1eadu. H\u1ecfi trung b\u00ecnh m\u1ed7i xe \u0111\u1ea9y chuy\u1ec3n \u0111\u01b0\u1ee3c bao nhi\u00eau ch\u1eadu hoa h\u1ed3ng?<\/p>\n

Gi\u1ea3i:<\/strong><\/p>\n

L\u1ea7n \u0111\u1ea7u, c\u1eeda h\u00e0ng chuy\u1ec3n \u0111\u01b0\u1ee3c s\u1ed1 ch\u1eadu hoa h\u1ed3ng l\u00e0: 3 x 16 = 48 (ch\u1eadu)<\/p>\n

L\u1ea7n sau, c\u1eeda h\u00e0ng chuy\u1ec3n \u0111\u01b0\u1ee3c s\u1ed1 ch\u1eadu hoa h\u1ed3ng l\u00e0: 5 x 24 = 120 (ch\u1eadu)<\/p>\n

Trung b\u00ecnh m\u1ed7i xe \u0111\u1ea9y chuy\u1ec3n \u0111\u01b0\u1ee3c s\u1ed1 ch\u1eadu hoa h\u1ed3ng l\u00e0: (48 + 120) : (3+5) = 21 (ch\u1eadu)<\/p>\n

V\u1eady trung b\u00ecnh m\u1ed7i xe \u0111\u1ea9y chuy\u1ec3n \u0111\u01b0\u1ee3c 21 ch\u1eadu hoa h\u1ed3ng.<\/p>\n

B\u00e0i to\u00e1n 7<\/h3>\n

\u0110\u1ec1 b\u00e0i:<\/strong>\u00a0L\u1ecd th\u1ee9 nh\u1ea5t c\u00f3 32 vi\u00ean bi, l\u1ecd th\u1ee9 hai c\u00f3 36 vi\u00ean bi. L\u1ecd th\u1ee9 ba ch\u1ee9a s\u1ed1 vi\u00ean bi nhi\u1ec1u h\u01a1n trung b\u00ecnh c\u1ed9ng c\u1ee7a hai l\u1ecd v\u1eeba r\u1ed3i 15 vi\u00ean bi. H\u1ecfi s\u1ed1 vi\u00ean bi c\u00f3 trong l\u1ecd th\u1ee9 3?<\/p>\n

Gi\u1ea3i:<\/strong><\/p>\n

S\u1ed1 trung b\u00ecnh c\u1ed9ng c\u1ee7a l\u1ecd \u0111\u1ef1ng vi\u00ean bi th\u1ee9 nh\u1ea5t v\u00e0 th\u1ee9 hai l\u00e0: (32 + 36) : 2 = 34 (vi\u00ean bi)<\/p>\n

S\u1ed1 vi\u00ean bi c\u00f3 trong l\u1ecd \u0111\u1ef1ng vi\u00ean bi th\u1ee9 ba l\u00e0: 34 + 15 = 49 (vi\u00ean bi)<\/p>\n

V\u1eady c\u00f3 49 vi\u00ean bi \u1edf trong l\u1ecd th\u1ee9 ba.<\/p>\n

\"cong-thuc-tinh-trung-binh-cong-la-gi-7\"<\/picture><\/p>\n

Cu\u1ed1i c\u00f9ng l\u00e0 m\u1ed9t s\u1ed1 b\u00e0i to\u00e1n n\u00e2ng cao, \u0111\u1ec3 c\u1ee7ng c\u1ed1 ki\u1ebfn th\u1ee9c cho ch\u00fang ta sau khi t\u00ecm hi\u1ec3u n\u1ed9i dung trung b\u00ecnh c\u1ed9ng l\u00e0 g\u00ec.<\/p>\n

B\u00e0i to\u00e1n 8<\/h3>\n

\u0110\u1ec1 b\u00e0i:<\/strong> C\u00f3 3 k\u1ec7 s\u00e1ch, trung b\u00ecnh m\u1ed7i k\u1ec7 x\u1ebfp \u0111\u01b0\u1ee3c 300 quy\u1ec3n s\u00e1ch, bi\u1ebft trung b\u00ecnh c\u1ed9ng c\u1ee7a k\u1ec7 th\u1ee9 nh\u1ea5t v\u00e0 k\u1ec7 th\u1ee9 hai l\u00e0 350. H\u1ecfi k\u1ec7 th\u1ee9 ba x\u1ebfp \u0111\u01b0\u1ee3c bao nhi\u00eau quy\u1ec3n s\u00e1ch?<\/p>\n

Gi\u1ea3i:\u00a0<\/strong><\/p>\n

T\u1ed5ng s\u1ed1 quy\u1ec3n s\u00e1ch c\u00f3 th\u1ec3 x\u1ebfp l\u00ean 3 k\u1ec7 l\u00e0: 3 x 300= 900 (quy\u1ec3n s\u00e1ch)<\/p>\n

T\u1ed5ng s\u1ed1 quy\u1ec3n s\u00e1ch x\u1ebfp tr\u00ean k\u1ec7 th\u1ee9 nh\u1ea5t v\u00e0 k\u1ec7 th\u1ee9 hai l\u00e0:\u00a02 x 350 = 700 (quy\u1ec3n s\u00e1ch)<\/p>\n

T\u1ed5ng s\u1ed1 quy\u1ec3n s\u00e1ch x\u1ebfp l\u00ean k\u1ec7 th\u1ee9 ba l\u00e0: 900 \u2013 700 = 200 (quy\u1ec3n)<\/p>\n

V\u1eady k\u1ec7 th\u1ee9 ba x\u1ebfp \u0111\u01b0\u1ee3c 200 quy\u1ec3n s\u00e1ch.<\/p>\n

B\u00e0i to\u00e1n 9<\/h3>\n

\u0110\u1ec1 b\u00e0i: <\/strong>M\u1ed9t gia \u0111\u00ecnh c\u00f3 5 ng\u01b0\u1eddi con. Trung b\u00ecnh c\u1ed9ng s\u1ed1 tu\u1ed5i c\u1ee7a 5 ng\u01b0\u1eddi con l\u00e0 11 tu\u1ed5i. N\u1ebfu t\u00ednh trung b\u00ecnh c\u1ed9ng c\u00f3 th\u00eam c\u1ea3 tu\u1ed5i c\u1ee7a ng\u01b0\u1eddi m\u1eb9 th\u00ec l\u00e0 14 tu\u1ed5i. V\u1eady ng\u01b0\u1eddi m\u1eb9 bao nhi\u00eau tu\u1ed5i?<\/p>\n

Gi\u1ea3i:\u00a0<\/strong><\/p>\n

T\u1ed5ng s\u1ed1 tu\u1ed5i c\u1ee7a ng\u01b0\u1eddi m\u1eb9 v\u00e0 5 ng\u01b0\u1eddi con l\u00e0: 14\u00a0 x 6 = 84 (tu\u1ed5i)<\/p>\n

T\u1ed5ng s\u1ed1 tu\u1ed5i c\u1ee7a 5 ng\u01b0\u1eddi con l\u00e0: 11 x 5 = 55 (tu\u1ed5i)<\/p>\n

S\u1ed1 tu\u1ed5i c\u1ee7a ng\u01b0\u1eddi m\u1eb9 l\u00e0: 84 – 55 = 29 (tu\u1ed5i)<\/p>\n

V\u1eady s\u1ed1 tu\u1ed5i c\u1ee7a ng\u01b0\u1eddi m\u1eb9 l\u00e0 29 tu\u1ed5i.<\/p>\n

B\u00e0i to\u00e1n 10<\/h3>\n

\u0110\u1ec1 b\u00e0i: <\/strong>Cho bi\u1ebft, trung b\u00ecnh c\u1ed9ng c\u1ee7a hai s\u1ed1 b\u1eb1ng 20. H\u00e3y t\u00ecm gi\u00e1 tr\u1ecb hai s\u1ed1, bi\u1ebft s\u1ed1 l\u1edbn g\u1ea5p ba s\u1ed1 b\u00e9.<\/p>\n

Gi\u1ea3i:<\/strong><\/p>\n

T\u1ed5ng c\u1ee7a hai s\u1ed1 c\u1ea7n t\u00ecm l\u00e0: 20 x 2 = 40<\/p>\n

S\u1ed1 b\u00e9 l\u00e0: 40 \/ 4 = 10<\/p>\n

S\u1ed1 l\u1edbn l\u00e0: 40 – 10 = 30<\/p>\n

V\u1eady s\u1ed1 l\u1edbn l\u00e0 30, s\u1ed1 b\u00e9 l\u00e0 10.<\/p>\n

\u1ee8ng d\u1ee5ng c\u1ee7a trung b\u00ecnh c\u1ed9ng l\u00e0 g\u00ec?<\/h2>\n

Trung b\u00ecnh c\u1ed9ng l\u00e0 m\u1ed9t c\u00f4ng c\u1ee5 to\u00e1n h\u1ecdc \u0111\u01a1n gi\u1ea3n nh\u01b0ng c\u00f3 nhi\u1ec1u \u1ee9ng d\u1ee5ng th\u1ef1c t\u1ebf quan tr\u1ecdng trong nhi\u1ec1u l\u0129nh v\u1ef1c kh\u00e1c nhau.<\/p>\n

T\u00ednh to\u00e1n gi\u00e1 tr\u1ecb trung b\u00ecnh<\/h3>\n

Trung b\u00ecnh c\u1ed9ng \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng \u0111\u1ec3 t\u00ednh to\u00e1n gi\u00e1 tr\u1ecb trung b\u00ecnh c\u1ee7a c\u00e1c d\u1eef li\u1ec7u trong nhi\u1ec1u l\u0129nh v\u1ef1c nh\u01b0 kinh t\u1ebf, x\u00e3 h\u1ed9i, khoa h\u1ecdc,…<\/p>\n

Kinh t\u1ebf:<\/strong> Trung b\u00ecnh c\u1ed9ng \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng \u0111\u1ec3 t\u00ednh to\u00e1n gi\u00e1 c\u1ea3 trung b\u00ecnh c\u1ee7a c\u00e1c m\u1eb7t h\u00e0ng, thu nh\u1eadp b\u00ecnh qu\u00e2n \u0111\u1ea7u ng\u01b0\u1eddi, t\u1ef7 l\u1ec7 th\u1ea5t nghi\u1ec7p,…<\/p>\n

X\u00e3 h\u1ed9i:<\/strong> Trung b\u00ecnh c\u1ed9ng \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng \u0111\u1ec3 t\u00ednh to\u00e1n \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a h\u1ecdc sinh, t\u1ef7 l\u1ec7 ng\u01b0\u1eddi bi\u1ebft ch\u1eef, tu\u1ed5i th\u1ecd trung b\u00ecnh,…<\/p>\n

Khoa h\u1ecdc:<\/strong> Trung b\u00ecnh c\u1ed9ng \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng \u0111\u1ec3 t\u00ednh to\u00e1n nhi\u1ec7t \u0111\u1ed9 trung b\u00ecnh, l\u01b0\u1ee3ng m\u01b0a trung b\u00ecnh, t\u1ed1c \u0111\u1ed9 trung b\u00ecnh,…<\/p>\n

\"cong-thuc-tinh-trung-binh-cong-la-gi-2\"<\/picture><\/p>\n

So s\u00e1nh c\u00e1c t\u1eadp h\u1ee3p s\u1ed1<\/h3>\n

Nh\u01b0 \u0111\u00e3 t\u00ecm hi\u1ec3u trong n\u1ed9i dung trung b\u00ecnh c\u1ed9ng l\u00e0 g\u00ec, trung b\u00ecnh c\u1ed9ng l\u00e0 gi\u00e1 tr\u1ecb \u0111\u1ea1i di\u1ec7n cho m\u1ee9c \u0111\u1ed9 t\u1eadp trung c\u1ee7a c\u00e1c s\u1ed1 trong m\u1ed9t t\u1eadp h\u1ee3p. N\u00f3 \u0111\u01b0\u1ee3c t\u00ednh b\u1eb1ng c\u00e1ch c\u1ed9ng t\u1ea5t c\u1ea3 c\u00e1c gi\u00e1 tr\u1ecb trong t\u1eadp h\u1ee3p v\u00e0 chia cho s\u1ed1 l\u01b0\u1ee3ng gi\u00e1 tr\u1ecb. Do \u0111\u00f3, trung b\u00ecnh c\u1ed9ng gi\u00fap ta so s\u00e1nh m\u1ee9c \u0111\u1ed9 t\u1eadp trung c\u1ee7a c\u00e1c s\u1ed1 trong c\u00e1c t\u1eadp h\u1ee3p kh\u00e1c nhau.<\/p>\n

Gi\u1ea3 s\u1eed ta c\u00f3 hai l\u1edbp h\u1ecdc A v\u00e0 B v\u1edbi \u0111i\u1ec3m trung b\u00ecnh m\u00f4n To\u00e1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 7,5 v\u00e0 8,2. Qua so s\u00e1nh, ta c\u00f3 th\u1ec3 nh\u1eadn th\u1ea5y \u0111i\u1ec3m trung b\u00ecnh m\u00f4n To\u00e1n c\u1ee7a l\u1edbp B cao h\u01a1n \u0111i\u1ec3m trung b\u00ecnh m\u00f4n To\u00e1n c\u1ee7a l\u1edbp A, th\u1ec3 hi\u1ec7n m\u1ee9c \u0111\u1ed9 t\u1eadp trung \u0111i\u1ec3m s\u1ed1 cao h\u01a1n c\u1ee7a l\u1edbp B.<\/p>\n

L\u01b0u \u00fd, vi\u1ec7c s\u1eed d\u1ee5ng trung b\u00ecnh c\u1ed9ng \u0111\u1ec3 so s\u00e1nh m\u1ee9c \u0111\u1ed9 t\u1eadp trung ch\u1ec9 hi\u1ec7u qu\u1ea3 khi c\u00e1c t\u1eadp h\u1ee3p s\u1ed1 c\u00f3 c\u00e1c gi\u00e1 tr\u1ecb g\u1ea7n nhau. N\u1ebfu c\u00e1c t\u1eadp h\u1ee3p s\u1ed1 c\u00f3 c\u00e1c gi\u00e1 tr\u1ecb kh\u00e1c bi\u1ec7t l\u1edbn, vi\u1ec7c so s\u00e1nh b\u1eb1ng trung b\u00ecnh c\u1ed9ng c\u00f3 th\u1ec3 kh\u00f4ng ph\u1ea3n \u00e1nh ch\u00ednh x\u00e1c m\u1ee9c \u0111\u1ed9 t\u1eadp trung c\u1ee7a c\u00e1c s\u1ed1 trong t\u1eadp h\u1ee3p.<\/p>\n

V\u00ed d\u1ee5:<\/strong> Gi\u1ea3 s\u1eed ta c\u00f3 hai t\u1eadp h\u1ee3p s\u1ed1 A = {1, 2, 3, 4, 5} v\u00e0 B = {1, 3, 10, 15, 20}.<\/p>\n