Q.A normal distribution has mean 100 and standard deviation 15. Find the two values of X between which the middle 95% of the distribution lies.

The 'middle 95%' is a symmetric interval around the mean, corresponding to the well-known $z=\pm1.96$ boundary; reverse the standardization formula for both boundaries. …

Step 1 — Identify the z-boundary for the middle 95%. If 95% of the area is in the middle, then $2.5\%$ is left in EACH tail (since the curve is symmetric). This is the standard, well-known boundary $z = \pm1.96$ — confirmed from the table since $\Phi(1.96)=0.9750$, meaning exactly $1-0.9750=0.0250=2.5\%$ lies above $z=1.96$, and by symmetry another 2.5% lies below $z=-1.96$, leaving $100\%-2.5\%-2.5\%=95\%$ in the middle.

Step 2 — Reverse the standardization formula for both boundaries.

$X_2 = \mu + 1.96\sigma = 100 + 1.96(15) = 100+29.4 = 129.4$ …

Using z = 2.00 (a commonly memorised rough figure) instead of the more precise z = 1.96 for exactly 95% — the two are close but not identical, and a Std-12 answer should use the table's stated val …