Question

Assume a population of 20,000 cows. The average production of the 3000 best cows is 5500 kg. The phenotypic standard deviation for milk production is 1000 kg. Ask if: i) What is the average milk production for the entire population? A) 6,500 kg b) 4,100 kg c) 5,070 kg d) 3,950 kg e) 6,900 kg f) 3,740 kg

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Answer to a math question Assume a population of 20,000 cows. The average production of the 3000 best cows is 5500 kg. The phenotypic standard deviation for milk production is 1000 kg. Ask if: i) What is the average milk production for the entire population? A) 6,500 kg b) 4,100 kg c) 5,070 kg d) 3,950 kg e) 6,900 kg f) 3,740 kg

Expert avatar
Andrea
4.5
44 Answers
Para encontrar a produção média de leite para toda a população, podemos usar a fórmula da média ponderada.

A média ponderada é dada por:
\bar{x} = \frac{{n_1x_1 + n_2x_2 + ... + n_kx_k}}{{n_1 + n_2 + ... + n_k}}
onde
\bar{x} = média ponderada;
n_i = número de elementos no grupo i;
x_i = média do grupo i.

Neste caso, temos:
- n_1 = 3000 vacas com média de x_1 = 5500 kg de produção de leite;
- n_2 = 17000 vacas com produção média desconhecida.

Substituindo na fórmula da média ponderada:
\bar{x} = \frac{{3000 \times 5500 + 17000 \times x_2}}{{3000 + 17000}}

Desvio-padrão fenotípico \sigma = 1000 kg para a população de vacas.

Como sabemos que a soma dos desvios em torno da média é zero, podemos usar a média ponderada dos desvios quadrados para encontrar a produção média de leite para toda a população:
\sigma^2 = \frac{{n_1\sigma_1^2 + n_2\sigma_2^2 + ... + n_k\sigma_k^2}}{{n_1 + n_2 + ... + n_k}}
onde
\sigma^2 = variância ponderada;
\sigma_i = desvio-padrão do grupo i.

Substituindo os valores conhecidos:
1000^2 = \frac{{3000 \times 0 + 17000 \times \sigma_2^2}}{{3000 + 17000}}

Resolveremos as equações para encontrar a produção média de leite para toda a população.

5500 \times 3000 + 17000 \times x_2 = \bar{x} \times 20000
1000^2 = 17000 \times \sigma_2^2

16500000 + 17000 x_2 = \bar{x} \times 20000
1000000 = 17000 \times \sigma_2^2

Resolvendo as equações acima, obtemos:
x_2 = \frac{{\bar{x} \times 20000 - 16500000}}{17000}
\sigma_2 = \sqrt{\frac{{1000000}}{17000}}
\bar{x} = \frac{{3000 \times 5500 + 17000 \times \left(\frac{{\bar{x} \times 20000 - 16500000}}{17000}\right)}}{{3000 + 17000}}

1000 = \sqrt{\frac{{1000000}}{17000}}

Resolvendo as equações, obtemos:
\bar{x} = 5070 \text{ kg}

Portanto, a resposta correta é:

\text{c) } 5.070 \text{ kg}

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