Suppose X ~ N(3 , 4). If sample size = 100, P(∑ Xi > 250) = ?. Draw
a figure.
Suppose "X\\sim N(\\mu, \\sigma^2)."
By the Central Limit Theorem for the random samples we take from the population, we can compute the mean of the sample means "\\mu_{\\bar{x}}=\\mu" and the standard deviation of the sample means "\\sigma_{\\bar{x}}=\\sigma\/\\sqrt{n}."
Then "\\bar{x}\\sim N(\\mu, \\sigma^2\/n)."
Given "\\mu=3, \\sigma^2=4, n=100."
"\\bar{x}\\sim N(3, 2^2\/100)."
"=1-P(\\bar{x}\\leq2.5)=1-P(Z\\leq\\dfrac{2.5-3}{2\/\\sqrt{100}})"
"=1-P(Z\\leq\\dfrac{2.5-3}{2\/\\sqrt{100}})=1-P(Z\\leq-2.5)"
"\\approx0.9938"
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