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2008Probability (Answer)

By Phillip Morgan,2014-06-11 17:51
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2008Probability (Answer)2008,08),2008)

    江西财经大学

    07;08学年第二学期期末考试试卷

    试卷代码,12104A 授课课时,64

    课程名称概率论Probability and Statistics 适用对象,2006级国际学院

    试卷命题人 刘满凤 试卷审核人

1. Fill in the blanks (3 points×5=15 points)

    (1)Suppose, then 0.3 P(A).,P(B).,P(AB).???040306PAB()

    (2)The continuous random variable X has exponential distribution, then the pdf of X is

    x?ex,0~ fx()0,0x?

    n12(3) A sampledrawn from the population. XX,X,X,?,XXN~(,)(,)12nin1in12222, then S(XX)E(SX);?(,;)i1n1i

    (4) A continues rvXhas a uniform distribution on interval, then the variance of Xis [0,]

    2 VX()/12

    (5) If random variable X has Binomial distribution , then the moment estimate of XBnp~(,)

    ˆparameter p is pXn/

    2. There are four choices in each question, but only one is correct. You should

    choose the correct one into the blank. (3 points×5=15 points)

    (1). For event A and B, if, then D( ABPAPB?????,0()1,0()1

    (A) (B) (C) (D) PAB(|)1PAB(|)1PBA(|)0PBA(|)0

    x(2).The discrete random variable X has a pmf , then PXxx()2,(1,2,...),01?????,,

    parameter is ( B )

    (A) 1/2 (B) 1/3 (C) 1/4 (D)ln2

    2(3). IfXfollows a normal distributionEX1EX4 is a simple sample (X,X,;,X)12n

    n1XXXdrawn from the population, then the distribution ofis C ( )in1i

    414313A( B( C( D( N(1,)N(,)N(,)N(1,)nnnnnn

    XY(4).Ifand are uncorrelated variables, then the incorrect formula is B (

    A( B( C( D( EXYEXEY()()()VXY()0cov(,)0XYVXYVXVY()()();?;

    (5).In tests about a normal population mean()the null hypothesis H:(( was rejected (00at significance level =0.01, if selected significance level =0.05the decision is C (

    A(H cannot be rejected B(either H cannot be rejected or H should be rejected 000

    C(H should be rejected D(H cannot be rejected and H should be rejected 000

    【第 1 4 页】

    3. (10 points) There are two children in one family. If any borne child is boy or girl is equal likely. Suppose

    A={ There is at least one boy in the family }

    B={If we know there is already one girl in the family, then there is exactly one boy and one girl in the family, }

    Determine the probability of event A,B

     Solution: PA()3/4

     PB()2/3

    4. (10 points) A transmitter is sending a message by using a binary code, namely, a sequence of 0s

    and 1s. Each transmitted bit (0 or 1) must pass through two relays to reach the receiver. At each relay, the probability is 0.20 that the bit sent will be different from the bit received (a reversal). Assume that the relays operate independently of one another.

    Transmitter?Relay 1?Relay 2?Receiver

    a. If a 1 is sent from the transmitter, what is the probability that a 1 is sent by both relays? b. If a 1 is sent from the transmitter, what is the probability that a 1 is received by the receiver? c. Suppose 70% of all bits sent from the transmitter are 1s. If a 1 received by the receiver, what is

    the probability that a 1 was sent?

    Solution: a. P???0.80.80.64

    b. P??;??0.80.80.20.20.68

    (0.80.80.20.2)0.7?;??P??0.832c. (0.80.80.20.2)0.7(0.80.20.20.8)0.3?;??;?;??

    5. (10 points) The reaction time (in seconds) to a certain stimulus is a continuous random variable with pdf

    31??;;,13x2 fx()2x

    0,otherwise?

    a. What is the probability that reaction time is at most 2.5 sec?

    b. Compute the expected reaction time.

    2.5319Solution: a. PXdx(2.5);???21210x

    3313b. EXxdx????ln32122x

    6. (10 points) The joint probability mass function for a discrete random vector X=(X, X) is given 12

    by

    XX;1?12,forX1,2;X1,212P(X,X) 1312

    0,otherwise?

    a. Determine the marginal probability mass function of X 1

    【第 2 4 页】

    b. What is? P(max(X,X))2122c. What is EX1

    Solution: a. the marginal probability mass function of X is 1

    X 1 2 1

    P 5/13 8/13

    b. P(max(X,X))/??221312

    58372c. EX??;??141131313

    7. (10 points) The random variable has a Poisson distribution, its pmf is X

    xe (),0,1,2,....???PXxx!x

    Please compute the maximum likelihood estimator of parameter.

    xxx;;;,,,n12,,,eeeSolution: the maximum likelihood function is (,,...,,)??fxxx12n!!!xxx12n

    n

     ln(,,...,,)()lnln(!!...!)fxxxxnxxx,,,?;;)1212nin1i

    n

    x)indfxxx(ln(,,...,,))112n1iˆ ?;?()0xn(??X)idn,,1i

    8. (10 points) A more extensive tabulation of t critical values than what appears in this book shows

    that for the t distribution with 20 df, the areas to the right of the values 0.687, 0.860, and 1.064 are

    0.25, 0.20, and 0.15, respectively. What is the confidence level for each of the following three

    confidence intervals for the mean of a normal population distribution? Which of the three (

    intervals would you recommend be used, and why? a. (0.687/21,1.725/21)xsxs;;

    b. (0.860/21,1.325/21)xsxs;;

    c. (1.064/21,1.064/21)xsxs;;

    Solution: Because , that is, the confidence ;??;?0.687(20),1.725(20),10.70ttSo0.250.05

    level of interval is 70%. (0.687/21,1.725/21)xsxs;;

    And other two confidence intervals have confidence level 70%.

    The width of is 2.412 (0.687/21,1.725/21)xsxs;;

    The width of is 2.185 (0.860/21,1.325/21)xsxs;;

    The width of is 2.128 (1.064/21,1.064/21)xsxs;;

    So I recommend the best interval is (1.064/21,1.064/21)xsxs;;

    【第 3 4 页】

9. (10 points) A certain type of automobile is known to sustain no visible damage 25% of the

    time in 10-mph crash tests. A modified bumper design has been proposed in an effort to increase

    this percentage. Let p denote the proportion of all 10-mph crashes with this new bumper that result

    in no visible damage. The test will be based on a sample of n=20.

    a. What are the appropriate null and alternative hypotheses?

    b. Let X=the number of crashes with no visible damage, and rejection region R={}. Please X~88

    calculate the probability of a type ?error and the probability of a type ?error when p=0.3.

    Solution: a. H:p=0.25 H: p>0.25 0a

     b. =P(type I error)=P(H is rejected when it is true) 0

     = PXwhenXBB(8~(20,0.25))1(7;20;0.25)10.8980.102~?;?;?

    P(type II error when p=0.3)=P(H0 is not rejected when it is false because p=0.3 (0.3)

     = PXwhenXBB(7~(20;0.3))(7;20;0.3)0.772;??

Table:

     ZZZ.05.025.01

    1.645 1.96 2.33

     ttttt.,02520.,02020.,1520.,1020.,0520

    0.687 0.860 1.064 1.325 1.725

Cumulative Binomial probabilities (n=20)

     P=0.2 P=0.25 P=0.3 P=0.4

    5 0.804 0.617 0.416 0.126

    6 0.913 0.786 0.608 0.250

    7 0.968 0.898 0.772 0.416

    8 0.990 0.959 0.887 0.596

    9 0.997 0.986 0.952 0.755

    【第 4 4 页】

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