2. A sprinters time for 100 metres may be recorded as 10.32 seconds but this means that his actual time was somewhere between 10.315 and 10.325 seconds. If e is the error in the recorded time, we can represent this will a pdf… 0 -0.005 0.005 e f(e) 100

3. This type of distribution is known as a uniform continuous distribution or a rectangular distribution. The properties of a uniform distribution are:  It takes the same value over the range in which it is defined.  It has no mode  It is symmetric about its mid-range value. Therefore the mean and median will both be equal to the mid-range value of the distribution.

4. If a continuous random variable, X, has a uniform distribution over the interval (α, β), then its probability density function will be given by The cumulative distribution function is given by

5. Proof of the variance result:

6. Example 1 The continuous random variable, X, is uniformly distributed over the interval (5, 12). Find: (a)E(X)(b)Var(X)(c)P(X < 10)

7. Example 2 A road is being repaired and temporary telephones are installed at intervals of 1km. A motorist breaks down on the road and can see 100 metres in either direction, but cannot see a phone. He tosses a coin to decide whether to walk forwards or backwards until he finds one. (a)Show that the distance he walks is uniformly distributed over the interval (100, 900) (b)Show that the mean distance he walks is half a kilometre (c)Find the probability that he has to walk less than a quarter of a kilometre. 1km = 1000m 100m (a) 100m 900m X ~ U(100,900)

8. Example 2 A road is being repaired and temporary telephones are installed at intervals of 1km. A motorist breaks down on the road and can see 100 metres in either direction, but cannot see a phone. He tosses a coin to decide whether to walk forwards or backwards until he finds one. (a)Show that the distance he walks is uniformly distributed over the interval (100, 900) (b)Show that the mean distance he walks is half a kilometre (c)Find the probability that he has to walk less than a quarter of a kilometre. (b) X ~ U(100,900) (c) P(X < 250)

9. Normal approximation to the binomial distribution Continuity correction Discrete random variables take only particular values, each with its own probability. Continuous random variables take values over an interval, and probabilities are defined for ranges of values rather than individual values. Consider the discrete variable X and specifically X = 4. 4 32 5 6 As a continuous variable, what would X = 4 actually mean? 3.5 < X < 4.5

10. Consider the discrete variable X and specifically X ≤ 5. 4 32 5 6 As a continuous variable, what would X ≤ 5 actually mean? X < 5.5 Consider the discrete variable X and specifically 3 < X ≤ 6. 4 32 5 6 As a continuous variable, what would X ≤ 5 actually mean? 3.5 < X < 6.5 …

11. Whenever you approximate a DISCRETE variable with a CONTINUOUS variable, you have to apply a continuity correction. Example 3 X is a discrete variable. Y is an approximation to X but is a continuous variable. Write down the probability you need to calculate for Y as the approximation for each of these probabilities for X. (a) P(X 12)(c) P(X ≤ 17)(d) P(12 < X ≤ 15) P(Y < 14.5) P(Y > 12.5) P(Y < 17.5) P(12.5 < Y < 15.5)

12. The parameters for the normal approximation If X ~ B(n, p) then E(X) = np and Var(X) = npq = np(1 – p) If n is large and p is close to 0.5, so that the distribution is nearly symmetrical, then you can use the normal distribution to approximate the binomial. The general rule is that the normal distribution can be used as an approximation when both np and np(1 – p) are > 5. X ~ B(n, p) ≈ Y ~ N(np, np(1 – p))

13. Example 4 If X ~ B(30, 0.4) calculate P(8 ≤ X ≤ 15). (a)from the tables of binomial probabilities (b)by using a normal approximation. (a)P(8 ≤ X ≤ 15) = P(X ≤ 15) – P(X ≤ 7) = 0.9029 – 0.0435 = 0.8594 (b)E(X) = np = 30 x 0.4 = 12 Var(X) = np(1 – p) = 30 x 0.4 x 0.6 = 7.2 X ~ B(30, 0.4) ≈ Y ~ N(12, 7.2) P(8 ≤ X ≤ 15) ≈ P(7.5 ≤ Y ≤ 15.5)

14. Example 5 An airline estimates that 7% of passengers who book seats on a flight do not turn up to take their seat. On one flight for which there are 185 available seats the airline sells 197 tickets. Find the probability that they will have to refuse boarding to any passengers holding a valid ticket for that flight. X ~ B(197, 0.07) ≈ Y ~ N(13.79, 12.8247) P(X ≤ 11) ≈ P(Y < 11.5)

15. Normal approximation to the Poisson distribution For a Poisson distribution with large λ (> 10) you would often use a normal distribution as an approximation, particularly when the probability of an interval is required, e.g. P(X ≥ 15) or P(6 < X < 14), since this will be a single calculation in a continuous distribution but will involve multiple calculations in a discrete distribution. X ~ Po(λ) ≈ Y ~ N(λ, λ)

16. Example 6 If X ~ Po(10), calculate P(8 ≤ X ≤ 15) (a)from the tables of Poisson probabilities (b)by using a normal approximation. (a)P(8 ≤ X ≤ 15) = P(X ≤ 15) – P(X ≤ 7) = 0.9513 – 0.2202 = 0.7311 (b)X ~ Po(10) ≈ Y ~ N(10, 10) P(8 ≤ X ≤ 15) ≈ P(7.5 ≤ Y ≤ 15.5)

17. Example 7 The demand for a particular spare part in a car accessory shop may be modelled by a Poisson distribution. On average the demand per week for that part is 5.5. (a)The shop has 7 in stock at the start of one week. What is the probability that they will not be able to supply everyone who asks for that part during the week? (b)The manager, who is going to be away for four weeks, wants to leave sufficient stock so that there is no more than a 5% probability of running out of any parts while he is away. How many of this particular spare part should he have in stock when he leaves? (a) X ~ Po(5.5) P(X > 7)= 1 – P(X ≤ 7) = 1 – 0.8095 = 0.1905

18. Example 7 The demand for a particular spare part in a car accessory shop may be modelled by a Poisson distribution. On average the demand per week for that part is 5.5. (a)The shop has 7 in stock at the start of one week. What is the probability that they will not be able to supply everyone who asks for that part during the week? (b)The manager, who is going to be away for four weeks, wants to leave sufficient stock so that there is no more than a 5% probability of running out of any parts while he is away. How many of this particular spare part should he have in stock when he leaves? (b) X ~ Po(22)≈ Y ~ N(22, 22) P(Z 0.95 P(Z > k) > 0.05 P(Z > 1.6449) = 0.05