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    Probability and Roravirus

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    (a) A student is taking a true-false exam with 20 questions. Assuming that this student guesses at all the 20 questions.
    (i) Determine the probability that this students gets either fourteen or sixteen answers correct.
    (ii) Approximate the probability obtained in part (i) by an area under a suitable normal curve.
    (b) (i) Under what circumstances do we have to make use of the continuity correction factor?
    (ii) If X is a random variable and n is the number of possible observations. Write all the five possible formulas we can apply continuity correction for x.
    (c) Acute rotavirus diarrhea is the leading cause of death among children under the age of 5, killing an estimated 4.5 million annually in development countries. Scientists from Finland and Belgium claim that a new oral vaccine is 80% effective against rotavirus diarrhea. Assuming that the claim is correct, find the probability that, out of 1500 cases, the vaccine will be effective in
    (i) Exactly 1225 cases
    (ii) At least 1175 cases
    (iii) Between 1150 and 1250 cases

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    NOTE: This posting was deserving 4 credits (NOT 2) -- you can see the effort has gone in it. Please raise a dummy posting of 2 credits in my favour. Thank you.

    (a) Probability of a correct answer, p = 1/2 == 0.5
    Probability of a wrong answer, q = 1 - 1/2 = 1/2 == 0.5

    Probability r out n correct answers = C(n,r)*p^r * q^ (n-r)

    Hence, probability of 14 correct answers,
    P(14) = C(20, 14) * (0.5)^14 * (0.5)^6
    = (20*19*18*17*16*15)/(1*2*3*4*5*6) * (0.5)^20 = 0.03696

    Probability of 16 correct answers,
    P(16) = C(20, 16) * (0.5)^16 * (0.5)^4
    = ...

    Solution Summary

    A fe problems related to probability. Also, a rotavirus diarrhea vaccine, statistics problem.