8.5, #5. Let’s say “most” means something like 65% of the plants. The cumulative distribution functions of the plants with fertilizer A, with no ferilizer, and with fertilizer B have value .65 when the heights of the plants are respectively .9, 1.1, and 1.3 meters. So fertilizer A stunts the growth of most plants, while fertilizer B promotes the growth of most plants.

8.5, #6. The density function for the duration of phones calls is 0.4e^{−0.4x}. The fraction of calls lasting between 1 and 2 minutes is

∫_{1}^{2} 0.4e^{−0.4x} dx = [−e^{−0.4x}]_{1}^{2} = 0.221 or 22.1%.

For calls lasting less than one minute, we integrate form 0 to 1 and get 33.0%. For calls lasting longer than 3 minutes, we integrate from 3 to infinity and get 30.1%. The cumulative distribution is

P (x) = ∫_{1}^{x} 0.4e^{−0.4u} du = 1 − e^{−0.4x} for x ≥ 0.

8.5, #8. The density function p(t) would be the % of times per minute the bus is t minutes late, where negative t would mean early arrival −t minutes ahead of schedule. So p(t) would have a graph of the following shape:

-3 -2 -1 0 1 2 3 4 5_{.}_{.}_{.}_{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} •

The area under the graph of p(t) between t = 2 and t = 4 represents the percentage of times the bus is late by 2 to 4 minutes. The cumulative distribution function P (t) has a graph of the following shape:

-3 -2 -1 0 1 2 3 4 5_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} •

The area under the graph of p(t) between t = 2 and t = 4 is P (4) − P (2) since P is an antiderivative of p(t). P (t) is concave-up when p(t) is increasing, concave-down when p(t) is decreasing. An inflection point of P (t) is a critical point of p(t). In this problem p(t) has only one critical point and P (t) has only one inflection point. P (t) is increasing at a faster and faster rate up to the inflection point, increasing at a slower and slower rate after the inflection point.

8.6, #1. The graph in figure 8.44 is linear from x = 2 to x = 6 and from x = 6 to x = 8. The area an be computed using the formula for the area of a trapezoid,

Area = (1/2)(0.08 + 0.24)(6 − 2) + (1/2)(0.24 + 0.12)(8 − 6) = 1.

The graph is a density function (for proportion of days with a given catch of fish), so the total area represents the total fraction, 1, of days considered.

8.6, #2. Since p is zero outside the interval [2, 8], the average daily catch is

\begin{aligned}∫_{2}^{8} xp(x) dx = & ∫_{2}^{6} x \{\frac{0.24 − 0.08}{6 − 2}(x − 2) + 0.08\} dx \\ & + ∫_{6}^{8} x \{\frac{0.12 − 0.24}{8 − 6}(x − 8) + 0.12\} dx = 5.253tons. \end{aligned}

The integrands are the linear functions whose graphs are pictured in figure 8.44.

8.6, #4. We are asked to approximate

\frac{1}{\sqrt{2π}} ∫_{−∞}^{∞} e^{−(x−15)^{2}/2} dx.

The difficulties are that the integral is improper and thta there is no elementary antiderivative. The integrand is largest near x = 15. We compute

\frac{1}{\sqrt{2π}} ∫_{10}^{20} e^{−(x−15)^{2}/2} dx.

by Simpson’s rule taking 4, 8, 16, and 32 subdivisions:

S_{4} = 0.9716, S_{8} =).9999, S_{16} = 1 − 6 · 10^{−7}, S_{32} = 1 − 6 · 10^{−7}.

To estimate the integral for 20 to infinity substitute u = x − 15 and note e^{−u^{2}/2} < e^{−2.5u} for u ≥ 5:

0 < \frac{1}{\sqrt{2π}} ∫_{20}^{∞} e^{−(x−15)^{2}/2} dx = \frac{1}{\sqrt{2π}} ∫_{5}^{∞} e^{−u^{2}/2} du < \frac{1}{\sqrt{2π}} ∫_{5}^{∞} e^{−2.5u} du = 6 · 10^{−7}.

Similarly

0 < \frac{1}{\sqrt{2π}} ∫_{−∞}^{10} e^{−(x−15)^{2}/2} dx < 6 · 10^{−7}.

So we estimate

1 − 6 · 10^{−7} < \frac{1}{\sqrt{2π}} ∫_{−∞}^{∞} e^{−(x−15)^{2}/2} dx, < 1 + 6 · 10^{−7}.