Homework 7 Solutions - Statistics Department
Homework 7 Solutions
1.
A Poisson process with [pic] accidents per week is equivalently a Poisson process with [pic] accidents per day. Thus, given that 7 accidents occurred in a particular week, the probability that exactly one accident occurred each day of that week is just
[pic].
For a Poisson process with parameter [pic], the waiting times between arrivals (in this case, accidents) are independent and exponentially distributed with mean [pic]. In this situation, that means the waiting times are exponentially distributed with mean 0.1 weeks between accidents, which corresponds to a mean of 16.8 hours between accidents. However, the distribution of an exponential random variable is heavily right-skewed – i.e., much of the mass of the pdf is concentrated below the mean. (See plot below for this exponential distribution over the interval [0, 1]) Hence, it is far more likely that waiting times less than 16.8 hours occur, and with shorter waiting times, it is difficult to have only 1 accident a day for 7 straight days.
[pic]
2. Ross 5.37 p. 251
a) If X is uniformly distributed over (-1, 1), then we have
[pic]
Hence, the probability that [pic] is
[pic].
b) To find the density function of [pic]:
[pic], where [pic].
Taking the derivative of this with respect to a gives us
[pic], where [pic].
In other words, [pic] is uniformly distributed on (0, 1).
3. Ross 5.41 p. 251
We have that [pic] is uniformly distributed on [pic]. If [pic], then to find the distribution of R, we just find the cdf for R as a function of x and then take the derivative with respect to x to obtain the pdf.
[pic]
4. Ross 6.6 p. 313
There are a total of [pic] different sequences in which the 5 transistors (2 defective) can be drawn from the bin. Each of these sequences is equally likely to occur, so the joint pdf of [pic] and [pic] is just
[pic]
5. Ross 6.11 p. 314
The customers can be viewed as draws from a multinomial distribution where [pic] is the probability of an “ordinary” purchaser, [pic] is the probability of a “plasma” purchaser, and [pic] is the probability of a browser. Hence, the probability that the store owner sells exactly 2 ordinary sets and 1 plasma set on a day when 5 customers enter his store is just
[pic].
6. Ross 6.10 p. 314
The joint pdf of X and Y is [pic], where x and y are both non-negative real numbers.
a) To find the probability that X is less than Y:
[pic]
b) For this part, note that the joint pdf of X and Y can be re-written as [pic]; that is, X and Y are independent. Hence, [pic] can be obtained by integrating [pic] and ignoring any mention of y (if X and Y were not independent, we would have to integrate [pic] over the full range of y and then integrate x over the interval (0, a)):
[pic].
7. Ross 6.14 p. 314
Let [pic] and [pic] denote respectively the locations of the ambulance and the accident at the moment the accident occurs. Then the distance is given by [pic], but this is equivalently the range of the two Uniform(0, L) random variables (i.e., [pic]). Using Equation (6.7), we get as the cdf of [pic]
[pic]
To obtain the pdf, take the derivative of the cdf with respect to a:
[pic].
8. Ross 6.15 p. 314
a) The joint density over the region R must integrate to 1, so we have
[pic],
where [pic] is the area of the region R; hence [pic].
b) Since the region R is a square with sides of length 2, [pic], and [pic] for [pic]. But this can be re-written as [pic], where [pic], and [pic]. In other words, X and Y are independent uniform random variables on the interval (-1, 1).
c) The probability that (X, Y) lies in the circle of radius 1 centered at the origin is just the area of the circle multiplied by 1/4:
[pic].
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