• 2022-05-27
    将一颗均匀的骰子连郑 10 次,求所得点数之和的方差.
  • 解 设 [tex=7.429x1.357]dGe2NwOUs7wEs1q/ypRUqdTCUAiWpRUK+LJ6ENi5IXg=[/tex] 为第 [tex=0.357x1.0]O88k7AtkDgTC9kv/8dY0lg==[/tex] 次掷骰子所出的点数,则[tex=13.857x4.214]NJBV0ERKMmdxR2RYoefQdAlTRqceQLtPrqdA1LqK4nBHy3WMkedB1ASxcE4yanOqkBdk3pxqy8w6sNVIKcTAAcIytibDCLKb6yy8AUjLt8A5wwKnj5jPQnVTapkuR4N4Uz+h1V/R+ysibh7TxZ5m926rN6yJW/1cuRViJNDVT1vaUDX7simSdGKBfalZ8GEzbnNvcXMojDHXF3+VdFy5FFBDtyYWHzf8U1C9zBrfYoo=[/tex]10 次捨股子的点数之和为[tex=4.5x3.429]gTBNZAWlGoXhmzs0Z81ZqoVQUYtsMr91xJGQ64RsJ7k=[/tex]易得[tex=17.929x3.429]uoVQ7klvXXMCcIqTYFS1ZJlGohaPAi5/Zq+hmbhqTTzyZL22IKCseDULWhSHtWaYZDb4fcW1jUGLMlfH9ymt9eiLaF3YFvedcU3QzQ6dvBfp2cDiQtoiFyJrM8B3HaK9[/tex]及[tex=16.786x3.429]6ROiOA/QWTyJ+ARZTxJJj4TWJvYH8vANclVjSirzDJTY9zulC7s0A1gRB7bpMQTibusDx7Y90hGoQKTilWe3ItQDdOktxPFY0gLI5t5SvVdBZALJdYJeM0iVqvkyw+Y5XCgdrSXJWvHt+tXQQEJeyQ==[/tex]又因 [tex=6.071x1.214]6m6IpLK9nxKlloS9uQjB0qJni044ihmKs30/YJo0lk0=[/tex] 相互独立,故[tex=7.857x3.429]w3ggUtrSFmnUXnFiZS5QdZBvAYM4YCDlVXcpj2kyrBVk/UaIN2AoX+bgTYJ5V+j1[/tex]其中[tex=19.143x4.214]+UKC6bniJvTTeNhVn6Y21GUqjjP5R0jhHeLt+dVrGLNaX6c7RYrKBbtSbRftdiVfrUPFqb/Mzk7aXkMAuKeoUNS4kyMiOXJXK9QvtaYceYA4QrLNo8i5g3LR+fgX1hY+ExFRU4xUgU2LGu6NthH44tSn74htND+vdc0KeIKz/qWUYYQbpxR3I1r3rPo28aANnNWCdOvU6oOxP7C5eCX9aMJizuI+qWOoGBp3r6KQDVG8JjFBKr+RsiXp3L0fk7hR[/tex]于是[tex=15.143x3.429]ZhIBTyq0Yso/IukDEIYe/L66jVj36TwPDsoAOpEqrufk5pFCeFkALFk17mEZ+Be3nJgIB22MQjpMSkNoxUbE+RgOlhVKmYJirzDmhRRJRZcoXSJz7ieXkhF8nROdtQ3r[/tex]

    内容

    • 0

      将一颗骰子抛掷两次,以[tex=1.214x1.214]AKRJ+piA0nf7C/6/dimpFw==[/tex]表示两次所得点数之和,求[tex=1.214x1.214]AKRJ+piA0nf7C/6/dimpFw==[/tex]的分布律.

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