Find P(X > 58.253) where X follows a normal distribution with mean 55 and standard deviation 10.

Step-by-step solution

1. 1

   Standardise the cut-off.

   $z=\frac{58.253-55}{10}=\frac{3.253}{10}=0.3253$

   The value is only about a third of a standard deviation above the mean, so the upper tail should come out a little under $0.5$.

2. 2

   Convert the "greater than" into a cumulative area. Standard normal tables give left-tail areas, so use the complement:

   $P(X>58.253)=P(Z>0.3253)=1-\Phi(0.3253)$

   Forgetting this subtraction — reporting $\Phi(z)$ itself — is the single most common error in right-tail problems.

3. 3

   Evaluate $\Phi$.

   $\Phi(0.3253)=0.62752$

   so

   $P(X>58.253)=1-0.62752=0.37248$

4. 4

   Round and state.

   $P(X>58.253)\approx 0.3725\quad\text{(about }37.25\%\text{)}$

5. 5

   Sanity-check against landmarks. At $z=0$ the right tail is exactly $0.5$, and at $z=0.5$ it is $0.3085$. Since $0.3253$ lies between $0$ and $0.5$, the answer must lie between $0.3085$ and $0.5$ — and $0.3725$ does $\checkmark$. Symmetrically, $P(X<51.747)$ equals the same $0.3725$.