Suppose that Y has a binomial distribution based on n trials and success probabilityp. Then ên = Y/n is an unbiased estimator of p. Use Theorem 9.3 to prove thatthe distribution of (ên − p)/√√ênân/n converges to a standard normal distribution. THEOREM 9.3Suppose that Un, has a distribution function that converges to a standard normaldistribution function as n → ∞. If W₁ converges in probability to 1, then thedistribution function of Un/W₁ converges to a standard normal distributionfunction.

To prove that the distribution of converges to a standard normal distribution, we'll use Theorem…

Answered: Suppose that Y has a binomial…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Let x be a random variable that represents the level of glucose in the blood (milligrams per deciliter of blood) after a 12-hour fast. Assume that for people under 50 years old, x has a distribution that is approximately normal, with mean μ = 93 and estimated standard deviation σ = 38. A test result x < 40 is an indication of severe excess insulin, and medication is usually prescribed. (a) What is the probability that, on a single test, x < 40? (Round your answer to four decimal places.)(b) Suppose a doctor uses the average x for two tests taken about a week apart. What can we say about the probability distribution of x? Hint: See Theorem 7.1. The probability distribution of x is approximately normal with μx = 93 and σx = 38.The probability distribution of x is approximately normal with μx = 93 and σx = 26.87. The probability distribution of x is approximately normal with μx = 93 and σx = 19.00.The probability distribution of x is not normal. What is the probability that x…
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Suppose that Y has a binomial distribution based on n trials and success probability p. Then pn = Y/n is an unbiased estimator of p. Use Theorem 9.3 to prove that the distribution of (Pn - p)/√ênân/n converges to a standard normal distribution.