Probability/Stats Worksheet 2 – Binomial/Geometric/Poisson/Gaussian Distributions

  1. For the following, provide (i) how you would define a random variable with the appropriate distribution and (ii) calculate the required probability.
  2. Example: Throwing 0 Heads in 4 throws of a fair coin.
  3. Throwing 2 Heads in 4 throws of a biased coin, with probability of Heads 0.4.
  4. Throwing a biased coin, with probability of Heads 0.7, until you see a Heads on the 5th throw.
  5. Receiving 5 visits to your website in the next minute, given that you see 4 visits on average.
  6. By finding a Z-Table (just Google it!) determine the probability of these particular IQ ranges, given that the mean IQ is 100 and the Standard Deviation is 15.
  7. An IQ less than 100.
  8. An IQ less than 115.
  9. An IQ less than 70 (your Z-Table might only have positive values of ; if so, think about the symmetry of the Normal Distribution).
  10. An IQ greater than 142.
  11. An IQ greater than 87.
  12. An IQ between 90 and 120.
  13. [Source: University interview] In an experiment to investigate animal behaviour rats have to make a choice between 4 doors of different colours. If they make the right choice they find food; if they make the wrong choice they get an electric shock. If an incorrect choice is made, the animal returns to its starting point and tries again, and this continues until a correct choice is made.
  14. If is the probability that a correct choice is made at the th attempt, find if:
  15. each door is equally likely to be chosen at each trial, and all trials are mutually independent;
  16. at each trial the rat chooses with equal probability between the doors which have not been tried previously, no choice ever being repeated;
  17. the rat never chooses the same door on two successive trials but otherwise chooses at random.
  18. Which strategy would be the best one for the rat to adopt?
  19. [Source: STEP] The number of texts that George receives on his mobile phone can be modelled by a Poisson random variable with mean texts per hour.
    Given that the probability George waits between 1 and 2 hours in the morning before he receives his first text is , show that:
    Hint: the volcano example in the slides might be helpful.
  20. (Difficult!) By noting that , prove that if (i.e. the geometric distribution with probability of success ), then .

ANSWERS

  1. Answers
  1. Answers:
  2. An IQ less than 100:
  3. An IQ less than 115:
  4. An IQ less than 70:
  5. An IQ greater than 142: 142 is s above the mean.
  6. An IQ greater than 87: By symmetry, this is the same as having an IQ less than 113. 113 is s away from the mean.
  7. An IQ between 90 and 120: The probability of an IQ less than 120 is 0.9082. Therefore the probability of an IQ between 100 and 120 is 0.9082-0.5=0.4082. The probability of an IQ greater than 90 is the same as the probability of an IQ less than 110, which is 0.7454. So the probability of an IQ between 90 and 100 is 0.7454-0.5=0.2454.
    Therefore the probability of an IQ between 90 and 120 is .
  8. Answers
  9. Answers:
  10. This is a geometric distribution.
  11. The most obvious thing to find is the expected value of each distribution, since this gives the expected number of times the rat will have to choose a door (obviously, the less the better!). Recall that if , then . Since the first strategy is , the expected number of doors is 4. For the second strategy, . For the last strategy, via some appropriate manipulation we get . The second strategy is therefore the best.
    Alternatively, you might look at the cumulative distribution function of each, but this would be difficult to calculate for all but the first strategy.
  12. Let be the number of texts George receives in the first hour of the morning and be the number he receives in the second hour.
    Then and , with and independent random variables.
    This we have
    Then
    Multiplying this last equation by gives , and a straightforward rearrangement yields our desired equation.
    Note that there is a distribution known as the exponential distribution which would directly allow us to find the probability of waiting a particular amount of time.
  13. We were able to move the differential function outside the summation, given that when we differentiate a sum, we differentiate each of the terms in turn, i.e. .
    We now conveniently have an infinite Geometry Series, where and . Thus using , we get . So continuing: