Q.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.]

Middle 60% $\Rightarrow z = \pm 0.84$ (area 0.30 each side); limits $= 52 \pm 0.84 \times 8 = (45.28,\ 58.72)$.

GSEB Class-12 Statistics, Normal Distribution:

Given $\mu = 52$, variance $= 64$, so $\sigma = \sqrt{64} = 8$.

The middle 60% of observations lie symmetrically about the mean, so 30% lie between the mean and each limit:

$\text{Area}(0 \text{ to } z) = 0.30$

From the standard normal table, the $z$ giving an area of $0.30$ is approximately $z = 0.84$. Hence the limits correspond to $z = \pm 0.84$.

Convert back to the original scale using $X = \mu \pm z\sigma$:

$X = 52 \pm 0.84 \times 8 = 52 \pm 6.72$

Lower limit $= 52 - 6.72 = 45.28$; upper limit $= 52 + 6.72 = 58.72$.

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