12. Exercises, Tutorials and Answers

Chapter 2: The Frequency Distribution

Chapter 2: The Relative Frequency Distribution

  • Exercise c02-01 - Constructing a Relative Frequency Distribution

  • Exercise c02-02 - Implement frequency distribution on Satellite Images

  • Question 1: The Relative Frequency Distribution or histogram for the whole image has three peaks, whilst those for sand and water do not. Why is this?
    Answer: The histograms for sand and water represent just the one cover type, whereas there are a number of cover types in the image, and some of these are sufficiently different to create a number of peaks in the histogram.

  • Question 2: The histogram for plantation does not either. Why not?
    Answer: The plantation sample includes a variation in the conditions that give differences in data values, leading to multiple peaks.

  • Question 3: What cover types are represented in the three peaks in the histogram for the whole image?
    Answer: There vary three peaks in the image histogram, of which one is represented by water and one by plantations. The third is represented by residential/commercial.

Chapter 2: Measures of Value

  • Exercise c03-01 - Using ILWIS for histogram calculation

  • Question 1: In the histogram above, the mean, median and mode are all marked as lines, with symbols (A), (B) and (C) against the lines. Which line represents the mean, the mode and the median?
    Answer: A=Mode, B= Mean, C = Median

  • Question 2: We have discussed here the sample mean, x ¯ . What is the population mean, µ and does the population mean vary from the sample mean and if so, why?
    Answer: The mean of the population is calculated using all members of the population. The population mean will normally be a little different to the sample mean, because not all members of the population are in the sample. The larger the sampling fraction, the closer the two will normally become.

  • Questions 3: If we call the differences between the individual observations, x 1 , x 2 , , x n and the mean, x ¯ , the residuals, then show, using a set of data, that the smallest sums of squares of the residuals occurs when using the mean value, that is
    Sof S smallest = { x 1 x ¯ } 2

    Answer: If x i = 23.5, 22.1, 24.2, 23.6, and 25.35 then the mean = 23.75 and the variances using different values for the mean are
    23.75 1.3925
    24 1.4706
    25 3.3406
    23 2.0956
    23.7 1.3956
    mean = 81.729, median = 128, mode = 83.

Chapter 2: The Multi-Dimensional Frequency Distribution

Chapter 2: Covariance and Correlation

Chapter 2: Probability

  • Question 1: What is the probability of at least one six?
    Answer: There are six combinations that contain one six, so the probability of throwing at least one six is 6·(1/36) or 1/6.

  • Question 2: What is the probability of getting 6 as the sum of the two dice?
    Answer: There are 5 combinations of the sum of the two dice that will give a six, so the answer is 5/36.

  • Question 3: If Pie in the Sky wins the first race, what is the probability that Doughboy will win the second?
    Answer: 0.35 since the probabilities of wining the second race are independent of what happened in the first race.

  • Question 4: What is the Probability that Raddish will win both races?
    Answer: 0.05*0.1=0.005

  • Question 5: What is the most likely combination to win?
    Answer: Pie in the Sky for the first and Doughboy for the second with a probability of 0.115.

  • Question 6: You want to make more than one bet on winning both races so that you have a probability of 25% or more that one of your combinations wins. What is the fewest number of combinations that you must select to reach 30%?
    Answer: 7.2 Three; Pie in the Sky and Doughboy = 0.1050, Atlantis Star and Doughboy = 0.0980 and Mums the Word and Doughboy = 0.0875

  • Question 7: In a deck of 52 cards, containing four suits, what is the probability of getting a King in the next dealt card?
    Answer: Pr{King} = 4/52 = 0.0769

  • Question 8: You are dealt a hand of 5 cards from a 52 card pack. What is the probability of getting an Ace?
    Answer: Pr{Ace in five cards} = 4/52+4/51+4/50+4/49+4/48 = 0.4

  • Question 9: Again you are dealt five cards. What is the probability of getting five spades?
    Answer: Pr{Five Spades} = (13/52) (12/51) (11/50) (10/49) (9/48) = 0.0000105

Chapter 2: Binomial Distribution

  • Question 1: What is the theoretical PDF for selecting a card at random of a particular number or rank (such as an Ace, 2, 3, ?, Jack, etc) from a pack of 52 cards? (Shuffle a pack of 52 cards and then select a single card. Repeat this process 30 times so as to derive a sample PDF to address the question)
    Answer: The theoretical distribution is the binomial distribution of the form f(x)=[ nx ]( 1/ 13 )x( 12 / 13 )( nx ) in which the [n x] has the n over the x as per the binomial and the x and (n - x) are power terms for (1/13) and (12/13) respectively, again as per the equation for the binomial function.

  • Question 2: If you shuffle the pack once and then select 30 cards at random from the pack, how does this PDF vary from the PDF derived in Question 1?
    Answer: The answer depends on the results the students get.

Chapter 2: Normal Distribution

  • Question 1: What is the probability that a value will occur within one standard deviation of the mean that is Pr{(-1≤z≤1)|N(0,1)}?
    Answer: Look up the Table for the Standard Normal PDF at 1.00 and you get 0.3413. This is from the centre out to one side of the distribution, so you need to double it, to get 0.6826.

  • Question 2: What is the probability that a value will occur in the range 8 to 12 in the Normal PDF, N(10,4)? Convert the x=8 and 12 to z values first.
    Answer: 0.6826

  • Question 3: What is the value of the probability,
    Pr{(10≤x≤14)|N(10,4)}?
    Answer: 0.4772

  • Question 4: What is the value of the probability,
    Pr{(9≤x≤14)|N(10,4)}?
    Answer: 0.3023

  • Question 5: What is the value of the probability,
    Pr{(11≤x≤15)|N(10,4)}?
    Answer: 0.6687

  • Question 6: What is the value of the probability,
    Pr{(x>12)|N(10,4)}?
    Answer: 0.1587