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Thread: Sta301 assignment 5, July, 2010

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    11 Sta301 assignment 5, July, 2010

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    Question 1: Marks: 2+3+5=10
    a) When you consider poisson distribution as the limiting form of the binomial distribution?

    b) The mean and standard deviation of the population is 30 and 5 respectively. The probability distribution of the parent population is unknown, find the mean and standard error of the sampling distribution of when n=50

    b) Ten vegetables cans, all of the same size, have lost their labels. It is known that 5 contain tomatoes and 5 contain corns. If 5 are selected at random, what is the probability that all contain tomatoes? What is the probability that 3 or more contain tomatoes?


    Question 2: Marks: 2+8=10
    a) Define sampling with replacement and sampling without replacement.

    b) A finite population consists of values 6, 6, 9, 15 and 18. Calculate the sample means for all possible random samples of size n=3, that can be drawn from this population without replacement. Make the sampling distribution of sample mean and find the mean and variance of this distribution.
    Question 3: Marks: 2+2+6=10
    a) Find the value of maximum ordinate of the standard normal curve correct to four decimal places.

    b) If Z is a standard normal random variable with mean 0 and variance 1, then find the Lower quartile.

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    c) Let be a random sample of size 3 from a population with mean
    Consider the following two estimators of the mean

    Which estimator should be preferred?


    PLZ GIVE ME SOLUTIONS

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    Mention the due date

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    15 july 2010

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    ab tu bs ek din reh ga hai ab tu koe upload kr de

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    plz anyone upload the solutin please

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    aik din bohot hota hia ji
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    Idea solution Please don't Copy

    b) A finite population consists of values 6, 6, 9, 15 and 18. Calculate the sample means for all possible random samples of size n=3, that can be drawn from this population without replacement. Make the sampling distribution of sample mean and find the mean and variance of this distribution.

    Solution :
    there are 5C3 (because order does not matter)ways of having a ***** of 3 from this finite population. So there are 10 groups and you have to find the mean of each of the 10 groups.
    6 6 9->7
    6 6 15->9
    6 6 18->10
    6 9 15->10
    6 9 15->10
    6 15 18->13
    6 15 18->13
    6 9 18->11
    6 9 18->11
    9 15 18->14
    Then it is asking you to find the average of the averages that you have found.
    (7+9+10+10+10+13+13+11+11+14)/10=10.8
    The variance is
    [(7-10.8)2+(9-10.8)2+(10-10.8)2+(10-10.8)2+(10-10.8)2+ (13-10.8)2+(13-10.8)2+(11-10.8)2+(11-10.8)2+(14-10.8)2]/(10-1)=4.4



    b) Ten vegetables cans, all of the same size, have lost their labels. It is known that 5 contain tomatoes and 5 contain corns. If 5 are selected at random, what is the probability that all contain tomatoes? What is the probability that 3 or more contain tomatoes?

    Solution:
    If all 5 are selected without replacement, there is only one way in which all 5 have tomatoes, C(5,5) = 1. Total number of ways of selecting 5 cans from 10 = C(10,5) = 180
    Hence, required probability of selecting all tomatoes = 1/180 ≈ 0. 005556


    The total number of ways of selecting 3 or more tomatoes is:
    C(5,3)*C(5,2) + C(5,4)*C(5,1) + C(5,5) = 126
    Total number of ways of selecting 5 cans from 10 = C(10,5) = 180
    Hence, required probability of selecting 3 or more tomatoes = 126/180 = 0.7
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    good work viki.. u done very good job

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