Homework #7 solutions

  • HW #7, due Friday 3/24:
  • 6.9 #3,4,5,6,8,16
  • 6.10 #3,6,8,9
  • 6.9#3

    a,i) III a,ii) VI
    b,i) I b,ii) V
    c,i) IV c,ii) II

    #4

    Density: something that peaks in the middle (class). Cumulative: something logistic-y, increasing slowly, then quickly, then slowly, up to 1.

    #5

    Density: start large and drop to a small value, then stay low. Cumulative: increase a lot first, then increase little, up to 1.

    #6

    Density: start large, go low, go high again (then go to zero). Cumulative: increase a lot, then stay nearly flat, then increase a lot, suddenly stopping at 1.

    #8

    The integral from A to B is the probability that somebody lives exactly that long (i.e. between A and B). Put differently, it's the probability that they die at a time in between A and B. So if A is zero, it's the probability that they die before time B. So P(t) = integral from 0 to t = the fraction that have died by time t. Note that it goes to 1 - everybody eventually dies.

    #16

    a) The function is 0, then some constant for x=0..5, then 0 again. We almost know it; all that's left is to determine the constant. All else we know is that the area is 1 - so the constant is 1/5. Therefore f(r)=1/5 for r=0..5, 0 otherwise.

    b) Integrating that, we get a function that's 0 for r<0, r/5 for r=0..5, and 1 for r>5.

    6.10#3

    a) Where does this cross y=1/2? Something like x=32.

    b) Call the cumulative function C. The density, C', is positive between 0 and 60 (C is always increasing, until it makes it up to 1). It's increasing where C is concave up, between 0 and 36 or so. It's decreasing where C is concave down, between 36 or so and 60. C gets most concave up at about 32, and most concave down at 45; these are the maxima and minima of C'.

    #6

    a) Typical exponential decrease.

    b) The indefinite integral is just -e^(-0.1t)+C, so the definite integral from 0 to x is 1-e^(-0.1x). For this to be 1/2 (where we get the median), we have

  • 1-e^(-0.1x)=1/2, so
  • e^(-0.1x)=1/2, so
  • -0.1x = ln 1/2, so
  • 0.1x = ln 2, so
  • x = 10 ln 2 or approximately 6.93 for those who needlessly insist on approximating.

    For the mean, compute the definite integral from 0 to 60 and divide by 60; this is ( 1-e^(-6) )/60, or really close to 1/60.

    The median time is how long to wait, such that half the times waited, you leave by that time. The mean time is the average over many times waited.

    #8

    a) The integral of this between 1.5 and 2, which is x^4/16 between 1.5 & 2, or 1-(1.5)^4/16 ~ 68%.

    b) The integral of [this times x], from 0 to 2, which is x^5/20 between 0 & 2, or 32/20 = 1.6 hours.

    c) The integral of this from 0 to T is x^4/16 between 0 and T, or T^4/16. This equals 1/2 at the median, so T^4 = 1/8, and T = the fourth root of 1/8, or about .59.

    #9

    a) 99-74=25%

    b) 50% of people made <12.6K/yr, so that's the median

    c) This is a cumulative, so you sketch the derivative, i.e. look at the differences. Note that the x scale has all these weird jumps (not all the same size). The biggets jump is around $8K. This means that the single income with the greatest number of people making it is $8K.