Dawson 201-MA2-DW tutoring

A density must be nonnegative and have total area 1. Use $t^2/9$ as an antiderivative of $2t/9$, and check it by differentiation. For an interval, evaluate the antiderivative at the upper bound and subtract its value at the lower bound. An exact time has an interval of zero width.

1. Read the density and its units

   The density is nonnegative on $[0,3]$, and it is zero elsewhere. Its graph over this interval is a rising straight line, from height 0 at $t=0$ to height $2/3$ per minute at $t=3$.

   For a short interval, density multiplied by interval width approximates probability. The units cancel: per minute multiplied by minutes gives a number without units. A definite integral adds these contributions across the interval.

2. Check that the whole area is one

   The power rule gives the antiderivative $t^2/9$, because its derivative is $2t/9$. The fundamental theorem of calculus then gives

   $$\int_0^3 \frac{2t}{9}\,dt=\left[\frac{t^2}{9}\right]_0^3=1.$$

   The bracket means “value at the upper bound minus value at the lower bound.” Nonnegative density and total area 1 together make this a valid continuous probability model. Geometrically, the same area is a triangle with base 3 minutes and height $2/3$ per minute: $\tfrac12(3)(2/3)=1$.

3. Translate the waiting-time interval into bounds

   Writing $\Pr$ for probability, the event “between 1 and 2 minutes” gives

   $$\Pr(1\le T\le2)=\int_1^2\frac{2t}{9}\,dt.$$

   Evaluate at both bounds: $2^2/9-1^2/9=4/9-1/9=1/3$. The probability is one third, or about 33.3%. Subtracting the lower-bound value removes the probability accumulated before 1 minute.

4. Check the answer using the graph

   The area under the density from 1 to 2 is a trapezoid of width 1 minute with heights $f(1)=2/9$ and $f(2)=4/9$ per minute. Its area is $\tfrac12(2/9+4/9)(1)=1/3$, confirming the integral.

   Multiplying only the right-end density by the width would give $4/9$, an overestimate here because the density rises across the interval. The exact trapezoid calculation works because this density is linear.

5. Separate a density value from a point probability

   At 2 minutes, the density is $f(2)=4/9$ per minute. The probability of exactly that time is $\Pr(T=2)=0$: integrating from 2 to 2 gives zero area.

   For this continuous model, including or excluding either interval endpoint does not change its probability. A recorded time rounded to “2 minutes” would describe a range of actual times; calculating that probability would require the rounding interval.