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Normal Approximations 3.6 Central Limit Theorem • Suppose X1, X2, . . . , Xn are independent measurements coming from a population with mean µ and variance σ 2. (i.e. the population does not have to be normal) • What is the sampling distribution of X? The density function for X is approximately normal with expected value µ and variance σ 2/n. • n should be ‘large’. Larger n is needed when the underlying population has more skewness, and smaller n is okay for ‘light-tailed’ symmetric populations. If the underlying is distribution is normal, then any value of n is fine. • Example. Under conditions of an accelerated life test, the time to failure of a transistor may be modelled as an exponential random variable with expected value 200 hours. ◦ Suppose a random sample of 3 such measurements is taken. What is the probability that the average of the measurements exceeds 225 hours? ◦ Now consider a random sample of 36 such measurements. What is the probability that the average of the measurements exceeds 225 hours? . P (X > 225) = P (Z > .75) = .227 ◦ Note that the standard deviation equals the expected value for exponential random variables. • Example. The lifetime of a type of battery is normally distributed with expected value 10 hours and standard deviation 3 hours. A package contains 4 batteries. When camping, I plan to use a flashlight that operates on 1 battery at a time for 35 hours. • Find the probability of running out of power early. P (X̄ < 35/4) = P (Z < −.833) = .203 • A more expensive battery has the same expected value, but a variance of 2.25 hours2. Find the probability of running out of power early with this brand. P (X̄ < 35/4) = P (Z < −1.6) = .0548 • Example. The ACME elevator company uses cables which will break when carrying more than 1000 pounds. 7 men board an elevator. If adult male weight is normally distributed with expected value 150 pounds and standard deviation 15 pounds, find the probability that the elevator cable will break. P (X̄ > 1000/7) = P (Z > −1.26) = .896 • Example. The breaking strength of a rivet has an expected value of 10000 psi and a variance of 250000 psi. The breaking strengths of 40 rivets are measured. Find the probability that the average of the measurements is between 9900 and 10200. . P (9900 < X̄ < 10200) = P (−1.26 < Z < 2.53) = .890 • Find the probability that 1 of the rivets has a breaking strength between 9900 and 10200? We don’t know the distribution of the measurements. 3.7 Approximations of Discrete Distributions • We can approximate a binomial random variable with large n by a normal random variable with mean np and variance np(1 − p). • We can approximate a Poisson random variable with large λ by a normal random variable with mean λ and variance λ. Examples • What is the approximate probability of observing more than 80 tails (in our coin drop experiment) in 100 independent trials, if the true probability of a tail is (a) 0.5? (b) 0.7? • Chocolate chips are randomly distributed in cookies with a mean of 15 per cookie. Find the approximate probability that a cookie would have more than 20 chocolate chips. • Exercise Suppose a random sample of 1500 voters is selected, and they are asked whether they support the Conservative Party. If the proportion supporting that party is .3 in the underlying population, what is the approximate probability that more than 33% of the sample says they support the Conservative Party?