I need help in the chapter-8 The Central Limit Theorem in Questions 8.5 vi) part(last). Also in questions 8.6, 8.7 and 8.8 please provide me with some detail on how P(...<Z<...) is calculated. Please help as soon as you can.
For 8.5(vi), we have to use a continuity correction to find an expression for X (where X is an integer) that is equivalent to 4<=10X<48. Since X is an integer, the values of 10X that lie within the range are 10,20,30,40, so X must be 1,2,3 or 4. So the probability required is P(0.5<X<4.5). [Remember that, using the continuity correction P(X=1) = P(0.5<X<1.5)] For 8.6, you are told to use a normal approximation, so you just find the mean and variance of the gamma distribution, and assume that X is approximately normally distributed with that mean and variance, ie (0.5, 0.01). We then find the probability as we would any other normal distribution, ie standardise so that P(X<0.8) = P(Z<[(0.8-0.5)/0.1]=P(Z<3). Qn 8.7 asks for the approximate probability, so you use the CLT. Even though the distribution of X is beta, the distribution of the sample mean Xbar will be approximately normally distributed with mean mu = 0.5, and variance = sigma^2/n = 0.0119/10. You then find the probability as usual. For 8.8, you are asked for an approximate probability so you use the normal approximation to the binomial. You find the mean and variance of the binomial and then use these as the mean and variance of the normal and find the required probability as usual. No continuity correction is required in 8.6 and 8.7 as the gamma and beta distributions are continuous but it is required in 8.8 as the binomial distribution is discrete.
