Please explain question 5.12 part two and three? How we got the value 0.392 and X has a N(mean, 1/25) and ho, 5 mean has a normal distribution? Thanks, Anjum
In part (ii) we're using the central limit theorem: \(\bar x \sim N(\mu , \sigma^2 / n)\) ie \(\frac {\bar x - \mu}{\sigma / \sqrt{n}} \sim N(0,1)\) 0.392 is 1.96/5
