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21. Figure 5-2 Consider the following picture of the density curve. Figure 5-3 The required area can be broken down into two parts. 21. 46. 1772. 4864. 6636. 5. The mean weight of 500 male students at a certain college is 151 lb and the standard deviation is 15 lb. Assuming the weights are normally distributed, ﬁnd how many students weigh (a) between 120 and 155 lb, (b) more than 185 lb. 5 lb. 5. 6000 This means that of the 500 male students polled, 60% of them weigh between 120 and 155 lb. 6000) = 300.

The distribution function F(x) has the following properties: 1. 2. , F(x) ≤ F(y) if x ≤ y]. lim F( x ) = 0; lim F( x ) = 1 3. , lim+ F( x + h) = F( x ) x→0 for all x]. x →−∞ x →∞ Distribution Functions for Discrete Random Variables The distribution function for a discrete random variable X can be obtained from its probability function by noting that, for all x in (-∞,∞), 0 −∞ < x < x1 x1 ≤ x < x 2 f ( x1 ) x 2 ≤ x < x3 F( x ) = f ( x1 ) + f ( x 2 ) M M f ( x1 ) + L f ( x n ) xn ≤ x < ∞ (4) It is clear that the probability function of a discrete random variable can be obtained from the distribution function noting that f ( x ) = F( x ) − lim− F(u) u→ x (5) CHAPTER 3: Discrete Random Variables 27 Expected Values A very important concept in probability and statistics is that of mathematical expectation, expected value, or brieﬂy the expectation, of a random variable.

180 3! (b) P(X > 2) = 1 − [ P( X = 0) + P( X = 1) + P( X = 2)] = 2 0 e −2 21 e −2 2 2 e −2 1− + + 1! 2! 0! 323 An exact evaluation of the probabilities using the binomial distribution would require much more labor. Relationships between Binomial and Normal Distributions If n is large and if neither p nor q is too close to zero, the binomial distribution can be closely approximated by a normal distribution with standardized random variable given by Z= X − np npq (9) Here X is the random variable giving the number of successes in n Bernoulli trials and p is the probability of success.