2. 1 P131, 10’s Solution Let X be the number of defective missiles that will not fire in one lot, then from the description of the problem, we have Hence, we have the mean and variance of X as follow So here Chebyshev inequality reads

3. 2 P131, 10’s Solution Which impies that the possibilities of the number of defective missiles vary corresponding to different k. For example, if k=1, X could be only 1 whose possibility will be at least zero, or X will take the value of 1 at the most possibilities; if k=2, X could be 0,1,2 whose possibility will be at least three fourths; For example, if k=5, X could be 0,1,2,3 whose possibility will be at least 24-25 th.

4. 3 Chapter 6 Some Continuous Probability Distributions Uniform Distribution Normal Distribution Gamma Distribution Exponential Distribution Chi-Squared Distribution

5. 4 6.1 Continuous Uniform Distribution Uniform Distribution The density function of the continuous random variable X on the interval [A, B] is f(x; A, B) = 1/(B – A), A  x  B = 0, elsewhere P(a  X  b) = (b – a) /(B – A)

6. 5 Theorem6.1 The mean and variance of the uniform distribution and The probability that X falls in an interval of fixed length within [A, B] is constant.

7. 6 Example 6.1, page 143 Suppose that a large conference room for a certain company can be reserved for no more than 4 hours. However, the use of the conference room is such that both long and short conferences occur quite often. In fact, it can be assumed that length of X of a conference has a uniform distribution on the interval [0, 4] What is probability density function? f(x)=1/4, 0  x  4 =0. elsewhere What is probability that any given conference lasts at least 3 hours? Example

8. 7 6.2 Normal Distribution The normal distribution is perhaps the most important distribution in statistical applications. (Central Limit Theorem) Many phenomena that occur in real world can be described approximately by normal distributions or normal curves. For examples: Heights, weights, Life time of a particular product, Average scores of the graduates from a university Measurements of manufactured parts, Errors in scientific measurements, The amount of rainfall in a city.

9. 8 Frequency distribution table N=110N=144

10. 9

11. 10

12. 11 The characteristics of frequency distribution y=f(x) ---- density function ( 密度函数 ) When n→∞, smooth line y=f(x)

13. 12 Definition of normal distribution A continuous random variable X having the bell-shaped distribution is called a normal random variable. The mathematical equation for the probability distribution of the normal variable depends upon the two parameters,  and , its mean and standard deviation. We denote the values of the density function of X by n(x; ,  ) n(x; ,  ) =

14. 13 The Properties of a Normal Curve The mode( 众数 ), which is the point on the horizontal axis where curve is a maximum, occurs at x = . The curve is symmetric about a vertical axis x = . The curve has its points of inflection ( 拐点 ) at x =   , is concave downward if  -  < x<  + , and concave upward otherwise. The normal curve approaches the horizontal axis asymptotically as we proceed in entire direction away from the mean. The total area under the curve and above the horizontal axis is equal to 1.

15. 14 The graph of a normal distribution density function y =f (x) μ μ+3σμ-3σ y x Bell-shaped σ< 1: steep σ>1: flat μ>0: right μ<0: left

16. 15 6.3 Area Under the Normal Curve The area under a density curve bounded by the two ordinates x = x 1 and x = x 2 equals the probability that the random variable X assumes a value between x 1 and x 2. P(x 1 < X < x 2 ) = n(x; ,  )dx = x-∞ +∞ 0≤F(x)≤1 x2x2 -∞ +∞ x1x1

17. 16 The difficulty in calculation of area Calculation for F(x) : 1)needs integration knowledge; 2)takes time. How to simplify the calculation of F(x)?

18. 17 Simplify to find out the value of F(x) calculate φ(z), and list them in a table (see p670-671 ， Table A.3). If X ～ N (μ,σ 2 ), then ～ N ( 0, 1 )

19. 18 Definition of Standard Normal Distribution The distribution of a normal random variable with mean zero and variance 1 is called a standard normal distribution. parameters: μ= 0, σ=1 (others are same)

20. 19 The relationship between normal distribution and standard normal distribution If X ～ N (μ,σ 2 ), then ～ N (0,1), Table A. 3 indicates the area under the standard curve corresponding to P(Z ≤ x).

21. 20 The relationship between normal distribution and standard normal distribution If X ～ N (μ,σ 2 ), then ～ N ( 0, 1 ) μ μ+3σ μ-3σ Normal distribution X f (x) 0 3 -3 Standard normal distribution Z Φ(z) Example

22. 21 1 ） P (z<-1)= φ(-1) = 2 ） P (z<1) = 1-φ(-1) = 3 ） P (-1<z<1)=1-2φ(-1) = 4 ） P (|z|>1) =2 φ(-1) = 0.1587 1-0.1587=0.8413 1-2*0.1587=0.6826 2*0.1587=0.3174 1 Z 0 Example a ： If z ～ N(0,1), find out the probability from the standard normal distribution table.

23. 22 Example b ： Estimating probability If 1000 body weights were drawn from a normal distribution with meanμ=70kg, sd σ=10kg. What is the portion of being less than 80kg? 70 80 1 Z X f (z) Φ(z) 0 Solution: LetThen, 0.8413 depends on table

24. 23 Example c: Estimating probability 7080 1 Z X f (z) Φ(z) 0 Solution: LetThen, If 1000 body weights were drawn from a normal distribution with meanμ=70kg, sd σ=10kg. What is the portion of being large than 80kg?

25. 24 Example d: Estimating probability Solution: LetThen, 70 80 1 Z X f (z) Φ(z) 0 50-2 -1 If 1000 body weights were drawn from a normal distribution with meanμ=70kg, sd σ=10kg. What is the portion of being large than 50kg and less than 80kg?

26. 25 Normal Probability 3  Rule a 1 2 3 P(|X-  |<a  ) 0.6826 0.9545 0.9973  －3－3 +3+3 正常区域 异常区域 异常区域 t X

27. 26 6.4 Applications of the Normal Distribution Example 6.13, page 155 The average grade for an exam is 74, and the standard deviation is 7. Assume the distribution of the grades of the exam is approximately normal, What percent of students scored above 85? P(X>85)=P((X-74)/7 >(85-74)/7)=P(Z>1.571) If 12% of students are given A’s, what is the lowest possible A? P(X>x)=12% Example

28. 27 6.5 Normal Approximation to the Binomial Theorem 6.2 If X is a binomial random variable with mean  = np and variance = np(1-p), then limiting form of the distribution of as n  , is the standard normal distribution n(z; 0, 1).

29. 28 Note The normal distribution with mean  = np and variance = np(1-p), provides: (a) A very accurate approximation to the binomial distribution when n is large and p is not extremely close to 0 or 1. (b) A good approximation to the binomial distribution even when n is small and p is reasonably close to 0.5. Often, we may use normal approximation to evaluate binomial probabilities if both np and n(1–p) are greater or equal to 5.

30. 29 n=5 p=0.3 n=10 p=0.3 The relationship between normal distribution and binominal distribution

31. 30 n=30 p=0.3

32. 31 Binomial,Poisson and Normal distribution Binomial Normal Poisson np>5 n(1-p)>5 n>50 λ>20 n>30 np<5

33. 32 Exercise 14, page 165 A commonly used practice of airline companies is to sell more tickets than actual seats to a particular flight because customers who buy ticket do not always show up for the flight. Suppose that the percentage of no shows at flight time is 2%. For a particular flight with 197 seats, a total of 200 tickets were sold. What is the probability that the airline overbooked this flight?

34. 33 Solution: X : the number of customers who will not show up for the flight among those 200 who bought the tickets X has a binomial distribution b(x; 200, 0.02) We need to find P(X  2). Since n = 200 is quite large, we use normal approximation.  = np = (200)(0.02) = 4  2 = np(1-p) = (200)(0.02)(0.98) =3.92,  = 1.98 z = = = -1.01 P(X  2) = P(Z  – 1.01)  0.1562.

35. 34 Example 2 Let X be a random variable whose distribution function is continuous and strictly increasing. Show that the random variable Y given by Y=F X (X) has a uniform distribution on [0,1]. Proof: for, we have Consequently, Note that for all random variables. However, the proof is more complicated when F X is not monotonic or continuous.

36. 35 Example 3 According to electrical circuit theory, the voltage drop across a resistor is related to the current flowing through the resistor by the equation V=IR, where R is the resistance level measured in ohms, I is the current in amperes, and V is the voltage in volts. In alternating current systems, the direction and magnitude of the current change in a cyclical pattern. Hence, if the resistance level is held constant, the voltage will also vary in a cyclical pattern. Consequently, measurements of the voltage will have a distribution that is symmetric about the mean. Suppose that the measured voltage in a certain electrical circuit has a normal distribution with mean 120 and standard deviation 2 and five measurements of the voltage are taken. Determine the probability that two of the measurements lie outside the range 118- 122.

37. 36 Solution Let X 1, X 2, X 3, X 4, X 5 be the voltage measurements. By assumption, X j ~ N(120, 2 2 ) for each j. Hence, Consequently, the probability that the jth measurement lies on outside the range 118-122 is 0.3174. Now let N be the number of voltage measurements that lie outside the range 118-122. Then N~B(5, 0.3174). Hence,

38. 37 Example 4 Measurements in science and engineering are always subject to error. Consider a scientist who is trying to determine the value of a particular physical constant whose true but unknown value is k. Suppose that the scientist’s measurement errors are normally distributed with mean 0 and standard deviation 1; that is, the scientist’s measurements are of the form k+E, where E ~ N(0, 1). Intuition suggests that the scientist can improve her estimate of the physical constant by taking a large number of independent measurements and averaging the results. Suppose that n independent measurements are made and let E 1,…, E n be the respective measurement errors. Then the estimate of the physical constant that the scientist obtains by averaging the measurements is

39. 38 Where. Hence, is the error in the scientist’s estimate of the physical constant. Hence,. Consequently, the distribution of becomes more concentrated around zero as n increases. This means that the scientist’s estimate of the physical constant improves with the number of observation taken, which is consistent with your intuition in the case.

40. 39 Example 5 Suppose that the scientist of the previous example would like her estimate to be within 0.01 of the true value with probability 95%. How many measurements must be taken? Continuing with the notation of the previous example, the scientist’s requirement is Since,where Z~N(0,1), this requirement is equivalent to From tables of the standard normal distribution,.Hence, the latter requirement is equivalent to That is, Consequently, if the scientists wishes the estimate to be within 0.01 of the true value with probability 95%, she must take at least 38,416 measurements!