Suppose x has a distribution with mu = 17 and sigma = 11. (a) If a random sample

Question

Suppose x has a distribution with mu = 17 and sigma = 11. (a) If a random sample of size n = 49 is drawn, find mu_x, sigma_x and P (17 lessthanorequalto x lessthanorequalto 19.) (Round sigma_x to two decimal places and the probability to four decimal places.) mu_x = sigma_x = P(17 lessthanorequalto x lessthanorequalto 19) = (b) If a random sample of size n = 57 is drawn, find mu_x, sigma_x = and p(17 lessthanorequalto x lessthanorequalto 19). (Round sigma_x to two decimal places and the probability to four decimal places) mu_x = sigma_x = p(17 lessthanorequalto x lessthanorequalto 19) = (c) Why should you expect the probability of part (b) to be higher than that of part (a)? The standard deviation of part (b) is part (a) because of the sample size. Therefore, the distribution about mu_x is .

Explanation / Answer

a) mean =17

for std error of mean =std deviation/(n)1/2 =1.57

P(17<X<19)=P(0<Z<1.2727)=0.8984-0.5 =0.3984

b)mean =17

std error =1.46

P(17<X<19)=P(0<Z<1.3727)=0.9151-0.5 =0.4151

c)std deviation of part b is smaller then part a becasue of larger sample size ........... abvout mu is narrower

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