8.6, #6. The fraction of the population making between $20000 and $50000 is P (50) − P (20) = 99 − 75 = 24%. The median income is the x for which P (x) = 50%, that is, x = $12600. The table for P (x) shows the fastest increase in the range 7.8 ≤ x ≤ 12.6, so the density function p(x) has its maximum in this interval, say at x = 9. This point would also be an inflection point of P (x).

8.6, #7. The probability density function for a gap in time of x seconds between cars on a certain freeway is

p(x) = \{ \begin{matrix}0 & x < 0 \\ ae^{−0.122x} & x ≥ 0. \end{matrix}

In order to have ∫_{0}^{∞} p(x) dx = 1, we must have a = 0.122. The cumulative distribution

P (t) = ∫_{0}^{t} 0.122e^{−0.122x} dx = 1 − e^{−.0122t}.

The median gap T satisfies the condition P (T) = 1/2, so T = ln 2/0.122 = 5.68sec. The mean is t̄= ∫_{0}^{∞} tp(t) dt = 8.2sec.

8.6, #9. A distribution of velocities is given by p(v) = av^{2}e^{−mv^{2}/(2kT)} for v ≥ 0 and p(v) = 0 for v < 0. The value of a is determined by the condition ∫_{0}^{∞} p(v) dv = 1. We use

the change of variables x = \sqrt{\frac{m}{kT}} v to simplify

∫_{0}^{∞} p(v) dv = a (\frac{kT}{m} )^{3/2} ∫_{0}^{∞} x^{2}e^{−x^{2}/2} dx.

Using integration by parts,

∫ x^{2}e^{−x^{2}/2} dx = −xe^{−x^{2}/2} + ∫ e^{−x^{2}/2} dx.

From the normal distribution we have

∫_{0}^{∞} x^{2}e^{−x^{2}/2} dx = \sqrt{\frac{2}{π}} ,

so ∫_{0}^{∞} p(v) dv = a (\frac{kT}{m} )^{3/2} \{0 + \sqrt{\frac{2}{π}} \}. Therefore

a = \sqrt{\frac{2}{π}} (\frac{m}{kT} )^{3/2} .

The mean speed is

v̄ = ∫_{0}^{∞} vp(v) dv = a ∫_{0}^{∞} v^{3}e^{−mv^{2}/(2kT)} dv = 2a (\frac{kT}{m} )^{2} = \sqrt{\frac{8}{π}} \sqrt{\frac{kT}{m}} = 1.596\sqrt{\frac{kT}{m}} ,

(using integration by parts).

The median speed V satisfies

1/2 = ∫_{0}^{V} p(v) dv = a ∫_{0}^{V} v^{2}e^{−v^{2}/(2kT)} dv.

Letting x = \sqrt{\frac{m}{kT}} v as above we get

1/2 = \sqrt{\frac{2}{π}} ∫_{0}^{V \sqrt{m/(kT)}} x^{2}e^{−x^{2}/2} dx.

This gives

∫_{0}^{V \sqrt{m/(kT)}} x^{2}e^{−x^{2}/2} dx = \sqrt{\frac{π}{8}} = 0.626657.

Using Simpson’s rule to approximate the integral we find

\begin{matrix}u & 1 & 2 & 1.5 & 1.6 & 1.53 & 1.538 \\ ∫_{0}^{u} x^{2}e^{−x^{2}/2} dx & 0.2491 & 0.9256 & 0.5989 & 0.6711 & 0.6207 & 0.62653 \end{matrix}

Hence V \sqrt{m/(kT)} = 1.538, so the median value V = 1.538\sqrt{\frac{kT}{m}} The maximum of p(v) occurs at a values of v for which p′(v) = 0. (This value of v is

called the mode.) Taking the derivative and solving for v we find v = \sqrt{\frac{2kT}{m}} = 1.414\sqrt{\frac{kT}{m}}.

The mode, median, and mean are in increasing order and are each proportional to \sqrt{T}.

8.Rev, #1. Let S be the solid obtained by rotating the given region around the x-axis, y = 0. Cuts perpendicular to the x-axis meet S in disks, the cut through x has radius \sqrt{x} and so has area πx. A Riemann sum approximating the volume os S is Σ_{i=1}^{n}πx_{i}Δx where

x_{i} = i/n. The volume is given by ∫_{0}^{1} πx dx = \frac{π}{2} .

8.Rev, #2. Rotating the same region around the horizontal line y = 1 the cuts are disks of radius 1 − \sqrt{x}, their areas are π(1 − \sqrt{x})^{2} and the volume is

π ∫_{0}^{1}(1 − \sqrt{x})^{2} dx = π ∫_{0}^{1} 1 − 2\sqrt{x} + x dx = \frac{π}{6} .

Rotating around the y-axis, x = 0, cuts perpendicular to the y-axis are disks of radius y^{2} and area πy^{4}. The volume is

∫_{0}^{1} πy^{4} dy = \frac{π}{5} .

8.Rev, #3. For the line y = ax A cut perpendicular to the x-axis at x meets the cone in a disk of radius bx/l which has area πb^{2}x^{2}/l^{2}. The volume is

π ∫_{0}^{l} b^{2}x^{2}/l^{2} dx = \frac{1}{3}b^{2}l.