Translate intervals and tails into normal probabilities

The mean is 50 and variance 8, so enter $\sigma=\sqrt 8$.

1. Press MODE and ensure STAT WIZARDS ON for guided entry.
2. Press 2 nd VARS, select 2:normalcdf, then ENTER.
3. Enter lower 48, upper 54, μ50 and $\sigma=\sqrt 8$; highlight Paste, then ENTER.
4. Press ENTER to calculate normalcdf(48,54,50,√8).

The displayed probability is 0.6816003607, giving 0.682 (3 s.f.). This is the area between 48 and 54, not the density at either endpoint.

Shade the right tail from 55. Use lower 55, upper $10^{99}$, μ50 and $\sigma=\sqrt 8$:

$\begin{aligned}&P(X>55)\approx\operatorname{normalcdf}(55,10^{99},50,\sqrt 8)\\ & =0.0385498886\approx\boxed{0.0385}.\end{aligned}$

The large upper bound simulates the infinite tail for this calculation. Recall the Teacher’s asymptote guidance: the density approaches $y=0$ without terminating at a finite x.

Shade everything left of 52. Use lower $-10^{99}$, upper 52, μ50 and $\sigma=\sqrt 8$:

$\begin{aligned}&P(X<52)\approx\operatorname{normalcdf}(-10^{99},52,50,\sqrt 8)\\ & =0.7602500133\approx\boxed{0.760}.\end{aligned}$

The lower bound simulates negative infinity. The source’s “graph below” remark refers to the left-tail graph actually printed above the input table.

Simplify before using the calculator:

$P(|X-51|<3)=P(-3<X-51<3)=P(48<X<54)=\boxed{0.682}\quad(3\text{ s.f.}).$

This is exactly the interval in(i). Its centre is 51, while the distribution mean is 50; do not replace it by a symmetric interval about 50.