**Homework assignment #3**

Complete exercises 2.7 and 2.14 in Bevington. The questions are reproduced below.

**2.7** At rush hour on
a typical day, 75% of the cars coming to an intersection turn right, and
25% turn left. (They must go either right or left). On a particular
day, 752 cars turn right and 283 turn left.

**a)** What is the predicted uncertainty
in the numbers of cars seen going left or right?

**b)** What is the probability
that this particular day was NOT a typical day. That is, what is the probability
that the number of cars turning right deviated from the average value by
the amount seen or greater?

* Hints: The situation in the problem is described by a binomial distribution, since there are a small number of outcomes (2). If you consider a right turn a "success", then p = 0.75. So, in part (a), you can use what you know about binomial distributions to estimate the expected uncertainty in the numbers of cars seen turning right or left (see Bevington chpt 2, or Taylor chpt 10). In part (b), you can use the fact that, although the parent distribution is a binomial one, there were a large number of observations (np>>1), so the binomial distribution is well-approximated by a Gaussian distribution. You may want to use table C.2 in Bevington (p. 253), or appendix A in Taylor (p . 245).

**2.14: ** This is a situation
frequently seen in counting experiments. It will apply directly to labs
7 and 9 and is also relevant for lab 11.

A problem arises when recording data with electronic counters in that the system may saturate when rates are very high, leading to a "dead time". For example, after a particle has passed through a detector, the equipment will be "dead" while the detector recovers and the electronics stores away the results. If a 2nd particle passes through the detector in this time period, it will not be counted.

**a) **Assume that a counter has
a dead time of 200 ns (200 x 10^{-9} sec) and is exposed to a beam
of 10^{6} particles per second (1 MHz). The mean number of
particles incident on the counter in a 200 ns time window is then m
= 0.2. From the Poisson probability for this process, find the efficiency
of the counter. That is, find the ratio of the mean number of particles
detected divided by the mean number of particles incident.

**b) ** Repeat the calculation
for beam rates of 0.02, 2, 4, 6, 8, 10, and 20 MHz, and make a graph of
counter efficiency as a function of beam rate.

Hint: The expression you derive should have the following limits: if the count rate is low, then the detector efficiency should be essentially 100% because the probability of another event occurring while the detector is dead will be low. Similary, if the count rate is much larger than 1 per 200 ns, the detector should be mostly "dead" and its efficiency will be very low.

**Answers:**

**2.7:** a) predicted uncertainty
= 13.9 cars

b) There is an 8.2% probability that the number of cars seen turning
right would be 752 or less or 800 or more, out of 1035 cars.

**2.14:** The expression for the
efficiency in some versions of Bevington solution is incorrect. It should
read eff = (1-e^{-}^{m})/m.
See the solution in Bevington for the derivation.