Statistics for laboratory scientists

Solutions for the homework problems for lecture 8

  1. Let p = Pr(male) = 105/205 = (approx) 0.512. Let X = number of males in 6 random newborns. The X ~ binomial(n=6, p=0.512). Let p(x) = Pr(X=x)

    x

    p(x)

    0

    1.3%

    1

    8.5%

    2

    22.3%

    3

    31.2%

    4

    24.6%

    5

    10.3%

    6

    1.8%

    In R, type round(dbinom(0:6, 6, 105/205)*100, 1)

  2. Let n = number of slides examined and X = number that are positive. If the sample is positive, then X ~ binomial(n, p=0.2).

    We seek n such that Pr(X = 0) <= 1%. But Pr(X=0) = (1-p)n = (0.8)n

    Thus, we wish to solve the following equation for n: (0.8)n <= 0.01.

    The solution: take logs.

    (0.8)n <= 0.01
    ln{(0.8)n} <= ln{0.01}
    n ln{(0.8)} <= ln{0.01}
    n (-0.223) <= (-4.605)
    n >= 20.6

    Thus, we must examine at least 21 slides.

    That is a lot of work! An improved method would be recommended.

    1. Let S = {fly has singed bristles} and W = {fly has white eyes}. S and W are independent, Pr(S) = 1/2, and Pr(W) = 1/4.

      Pr(S and W) = Pr(S) Pr(W) = (1/2) × (1/4) = 1/8 = (approx) 13%.

    2. Let Wi = {fly i has white eyes}. The Wi are independent, and Pr(Wi) = 1/4.

      Pr(all four have white eyes) = Pr(W1 and W2 and W3 and W4) = (1/4)4 = (approx) 4/1000.

    3. First, note that Pr(a fly has neither white eyes nor singed bristles) = Pr(not W and not S) = Pr(not W) × Pr(not S) = (3/4) × (1/2) = 3/8.

      Pr(none of four flies have either white eyes or singed bristles) = (3/8)4 = (approx) 2%.

    4. Pr(at least one of two flies has white eyes or singed bristles or both) = 1 - Pr(neither has white eyes or singed bristles or both) = 1 - Pr(both are not W and not S) = 1 - (3/8)2 = (approx) 86%.

  4. Pr(exactly 50 heads in 100 tosses) = (100 choose 50) (0.5)100 = (approx) 8%.

    Pr(exactly 3 heads in 10 tosses) = (10 choose 3) (0.5)10 = (approx) 12%.

    Thus, the latter is more likely.

    You can use the following R code, if your hand calculator is not sufficiently advanced: dbinom(50,100,0.5) and dbinom(3,10,0.5)

  5. Pr(getting a double-six) = 1/36. Thus, if we let X = number of double-sixes in 36 rolls of a pair of fair, six-sided dice, X ~ binomial(n=36, p=1/36).

    1. X ~ binomial(n=36, p=1/36)

    2. Pr(X=2) = (36 choose 2) (1/36)2 (35/36)34 = (approx) 19%

    3. E(X) = 36 × (1/36) = 1

    4. SD(X) = [36 × (1/36) × (35/36)](1/2) = (approx) 0.99

    5. Pr(X > 2) = 1 - Pr(X=0) - Pr(X=1) - Pr(X=2) = (approx) 8%.

    Parts (a) and (d) would be a lot easier on the computer: dbinom(2, 36, 1/36) and 1-pbinom(2,36,1/36).

  6. X follows a Poisson(lambda=2) distribution.

    1. E(X) = 2.

    2. SD(X) = sqrt(2) = (approx) 1.4.

    3. Pr(X = 0) = e-2 = (approx) 14%.

    4. Pr(X = 5) = e-2 25 / 5! = (approx) 3.6%.

    5. Pr(X > 2) = 1 - Pr(X = 0) - Pr(X = 1) - Pr(X = 2) = 1 - e-2 - e-2 2 - e-2 22 / 2! = (approx) 32%.

  7. E(Y) = 30 and SD(Y) = 5.

    1. Z = (Y - 30)/5. E(Z)=0 and SD(Z)=1.

    2. X = - Y. E(X) = -E(Y) = -30 and SD(X) = SD(Y) = 5.

    3. R = 5 + Y/3. E(R) = 5 + E(Y)/3 = 15 and SD(R) = SD(Y)/3 = (approx) 1.67.

  8. U ~ uniform(5, 10).

    1. E(U) = (5+10)/2 = 7.5

    2. Pr(U = 6) = 0 (U is a continuous random variable.)

    3. Pr(U > 6) = 4/5 = 80%

    4. Pr(7 < U < 9) = 2/5 = 40%

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Last modified: Wed Feb 22 09:44:17 EST 2006