Statistical Methods Assignment Help

Questions And Solutions

1. (based on K.Aho, 2014, p.57) A gill net survey conducted in the 1990s in a certain lake found that 55% of fish in the lake were cutthroat trout. Last year catch and release data from anglers provide species information for a total of 12 fish caught in the lake.

1.What is the probability that a) All 12 fish will be cutthroat trout 0.00077 b) Exactly 5 fish will be cutthroat trout 0.148945 c) Less than 5 fish will be cutthroat trout 0.11174 d) Five of more fish will be cutthroat trout 0.88826

2. a) Let’s say that actually only 1 of the caught fish was the cutthroat trout. What is the probability of this outcome given the 1990s fish proportions?

0.001011 b) This is a very small data set, so one should be very careful with making general conclusions from it. Still, what your answer in 2a might suggest about cutthroats dynamic in this lake?

It suggests that getting a single fish which is cut trout is very less probable

3. What is the average number of cutthroat trout that would be expected in the last year catch given the 1990s fish proportion? 6.6

4. What is the standard deviation of cutthroat trout that would be expected in the last year catch given the 1990s fish proportion? 1.723369

2. Let y be a Poisson random variable. Find the following values

a) P(y=3), given m=2.4 0.778722

b) P(y<=1), given m=2.3 0.3308542

c) P(y>1), given m=2.2 0.6454299

d) P(y<=5), given m=1.8 0.989622

3. An entomologist recorded the numbers of insect of a certain species caught from 20 insect traps within the study area. The table below shows the numbers of insects per trap that were observed (observed number) and the number of traps in which the certain number of insects was found (frequency). Calculate the mean and the variance for the data set. Put the answers ion the box below along with brief show of the work.

Mean: 1.15

Observed

Brief illustration of how you did the calculations: We calculated the sum of the product of the each observed number with frequency and then divided it total frequency. Variance: 4.0275

Brief illustration of how you did the calculations:

We calculated the sum of the product of the each squared observed number with frequency and then divided it total frequency and then subtracted the square of the mean from that

4) Find the area under the normal curve between these values. You can use either Table 1 from the text book, R, or SAS.

a) z=0.4 and z=1.3 0.2477778

b) z=-1.1 and z=0.5 0.5557964

c) z=-1.4 and z=-0.3 0.3013319

5) In a certain population of lizards, the length of individual animal is approximately normally distributed with mean of 74.0 mm and standard deviation of 5.5 mm. What is the probability that if you randomly select a lizard from that population its length will be (express the answer in %):

a) greater than 80 mm 0.8623436

b) between 67 and 60 mm 0.09610064

c) less than 70 mm