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《概率论》期末考试试题A卷和答案.docx

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《概率论》期末考试试题A卷和答案

一、选择题(每题3分,共15分)

1.设事件\(A\)与\(B\)互不相容,且\(P(A)=0.3\),\(P(B)=0.4\),则\(P(\overline{A}\cap\overline{B})=(\)\)

A.\(0.3\)

B.\(0.4\)

C.\(0.36\)

D.\(0.7\)

答案:A

详细解答:因为\(A\)与\(B\)互不相容,所以\(A\capB=\varnothing\),根据摩根定律\(\overline{A}\cap\overline{B}=\overline{A\cupB}\)。又\(P(A\cupB)=P(A)+P(B)=0.3+0.4=0.7\),所以\(P(\overline{A}\cap\overline{B})=P(\overline{A\cupB})=1P(A\cupB)=10.7=0.3\)。

2.设随机变量\(X\)服从参数为\(\lambda\)的泊松分布,且\(P(X=1)=P(X=2)\),则\(\lambda=(\)\)

A.\(1\)

B.\(2\)

C.\(3\)

D.\(4\)

答案:B

详细解答:泊松分布的概率公式为\(P(X=k)=\frac{\lambda^{k}e^{\lambda}}{k!}\),\(k=0,1,2,\cdots\)。已知\(P(X=1)=P(X=2)\),即\(\frac{\lambda^{1}e^{\lambda}}{1!}=\frac{\lambda^{2}e^{\lambda}}{2!}\),因为\(e^{\lambda}\neq0\),两边同时约去\(e^{\lambda}\),得到\(\lambda=\frac{\lambda^{2}}{2}\),由于\(\lambda0\),解得\(\lambda=2\)。

3.设随机变量\(X\)的概率密度为\(f(x)=\begin{cases}2x,0x1\\0,其他\end{cases}\),则\(P(X\leqslant0.5)=(\)\)

A.\(0.25\)

B.\(0.5\)

C.\(0.75\)

D.\(1\)

答案:A

详细解答:根据概率密度函数求概率公式\(P(X\leqslanta)=\int_{\infty}^{a}f(x)dx\),则\(P(X\leqslant0.5)=\int_{0}^{0.5}2xdx=x^{2}\big|_{0}^{0.5}=0.5^{2}0^{2}=0.25\)。

4.设\(X\simN(1,4)\),\(\varPhi(x)\)为标准正态分布的分布函数,则\(P(1X3)=(\)\)

A.\(\varPhi(1)\varPhi(1)\)

B.\(\varPhi(2)\varPhi(2)\)

C.\(\varPhi(\frac{31}{2})\varPhi(\frac{11}{2})\)

D.\(\varPhi(3)\varPhi(1)\)

答案:C

详细解答:若\(X\simN(\mu,\sigma^{2})\),则\(Z=\frac{X\mu}{\sigma}\simN(0,1)\)。已知\(X\simN(1,4)\),即\(\mu=1\),\(\sigma=2\),那么\(P(1X3)=P(\frac{11}{2}\frac{X1}{2}\frac{31}{2})=P(1Z1)=\varPhi(\frac{31}{2})\varPhi(\frac{11}{2})\)。

5.设\(X_1,X_2,\cdots,X_n\)是来自总体\(X\simN(\mu,\sigma^{2})\)的样本,\(\overline{X}=\frac{1}{n}\sum_{i=1}^{n}X_i\),则\(D(\overline{X})=(\)\)

A.\(\sigma^{2}\)

B.\(\frac{\sigma^{2}}{n}\)

C.\(n\sigma^{2}\)

D.\(\frac{\sigma^{2}}{n^{2}}\)

答案:B

详细解答:因为\(X_1,X_2,\cdots,X_n\)相互独立且都服从\(N(\mu,\sigma^{2})\),根据方差的性质\(D(aY)=a^{2}D(Y)\)和\(D(X_1+X_2+\cdots+X_n)=D(X_1)+D(X_2)+\cdots+D(X_n)\)(\(X_i\)相互独立),则\(D(\overline{X})=D(\frac{1}{n}\sum_{i

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