Question
A normal distribution has mean 52 and variance 64. Obtain estimated limits which include exactly middle 60 % of the observations.
[Note : Blind students should define standard normal variable and write its probability function.]

Answer

Middle 60% means 30% area (0.3000) on each side of mean.
From Z-table, area 0.3000 corresponds to $z \approx \pm 0.84$.
$x = \mu \pm z\sigma = 52 \pm 0.84(8) = 52 \pm 6.72$.
Limits : (45.28, 58.72).

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