Exams Joint Entrance Examination Mathematics Statistics Question 40 Question 40 of 208 Hard MCQ
Let the mean and standard deviation of marks of class A of 100 students be respectively 40 and α ( > \alpha(> α ( >
0 ), and the mean and standard deviation of marks of class B B B of n n n students be respectively 55 and 30
− α -\alpha − α . If the mean and ...
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A B A + B x ⃗ 1 = 40 x ⃗ 2 = 55 x ⃗ = 50 σ 1 = α σ 2 = 30 − α σ 2 = 350 n 1 = 100 n 2 = n 100 + n x ⃗ = 100 × 40 + 55 n 100 + n \begin{gathered} \, A \quad B \quad A+B \\ {\vec{x}}_{1}=40 \quad {\vec{x}}_{2}=55 \quad \vec{x}=50 \\ {\sigma }_{1}=\alpha \quad {\sigma }_{2}=30-\alpha \quad {\sigma }^{2}=350 \\ {n}_{1}=100 \quad {n}_{2}=n \quad 100+n \\ \vec{x}=\frac{100\times 40+55n}{100+n} \quad \end{gathered} A B A + B x 1 = 40 x 2 = 55 x = 50 σ 1 = α σ 2 = 30 − α σ 2 = 350 n 1 = 100 n 2 = n 100 + n x = 100 + n 100 × 40 + 55 n
5000 + 50 n = 4000 + 55 n 1000 = 5 n n = 200 σ 2 1 = ∑ x 2 i 100 − 40 2 σ 2 2 = ∑ x 2 j 100 − 55 2 350 = σ 2 = ∑ x i 2 + ∑ x 2 j 300 − ( x ⃗ ) 2 350 = ( 1600 + α 2 ) × 100 + [ ( 30 − α ) 2 + 3025 ] × 200 300 − ( 50 ) 2 2850 × 3 = α 2 + 2 ( 30 − α ) 2 + 1600 + 6050 8550 = α 2 + 2 ( 30 − α ) 2 + 7650 α 2 + 2 ( 30 − α ) 2 = 900 α 2 − 40 α + 300 = 0 α = 10 , 30 σ 2 1 + σ 2 2 = 10 2 + 20 2 = 500 \begin{gathered} 5000+50n=4000+55n \\ 1000=5n \\ n=200 \\ {\sigma }_{2}^{1}=\frac{\sum {x}_{2}^{i}}{100}-{40}^{2} \\ {\sigma }_{2}^{2}=\frac{\sum {x}_{2}^{j}}{100}-{55}^{2} \\ 350={\sigma }^{2}=\frac{\sum {x}_{i}{}^{2}+\sum {x}_{2}^{j}}{300}-(\vec{x}{)}^{2} \\ 350=\frac{(1600+{\alpha }^{2})\times 100+[(30-\alpha {)}^{2}+3025]\times 200}{300}-(50{)}^{2} \\ 2850\times 3={\alpha }^{2}+2(30-\alpha {)}^{2}+1600+6050 \\ 8550={\alpha }^{2}+2(30-\alpha {)}^{2}+7650 \\ {\alpha }^{2}+2(30-\alpha {)}^{2}=900 \\ {\alpha }^{2}-40\alpha +300=0 \\ \alpha =10,30 \\ {\sigma }_{2}^{1}+{\sigma }_{2}^{2}={10}^{2}+{20}^{2}=500 \end{gathered} 5000 + 50 n = 4000 + 55 n 1000 = 5 n n = 200 σ 2 1 = 100 ∑ x 2 i − 40 2 σ 2 2 = 100 ∑ x 2 j − 55 2 350 = σ 2 = 300 ∑ x i 2 + ∑ x 2 j − ( x ) 2 350 = 300 ( 1600 + α 2 ) × 100 + [( 30 − α ) 2 + 3025 ] × 200 − ( 50 ) 2 2850 × 3 = α 2 + 2 ( 30 − α ) 2 + 1600 + 6050 8550 = α 2 + 2 ( 30 − α ) 2 + 7650 α 2 + 2 ( 30 − α ) 2 = 900 α 2 − 40 α + 300 = 0 α = 10 , 30 σ 2 1 + σ 2 2 = 10 2 + 20 2 = 500 Answer the question to unlock