So far you went from a value to an area. Often the question runs the other way: "How long must a battery last to be in the top 5%?" Now you start with an area, find the z-score that cuts it off, and then turn the z-score back into a value, the reverse of finding areas.
Idea clave
To find the value with a given area to its left: (1) look up the z with that left area (inverse normal); (2) undo the z-score: $x = \mu + z\sigma$. For "top $k\%$," use left area $1 - k/100$.
Por qué funciona
A z-score is a position measured in standard deviations. Reading the table backward tells you that position for the area you want. Since $z = (x - \mu)/\sigma$, solving for $x$ gives $x = \mu + z\sigma$: start at the mean and walk $z$ standard deviations.
Ejemplo resuelto: A percentile
Battery lifetimes are normal with mean 500 hours and SD 50. Find the 90th percentile. (The z with 90% to its left is $1.28$.)
- The 90th percentile has 90% of the area to its left, so $z = 1.28$.
- $$x = 500 + 1.28(50) = 500 + 64 = 564$$
Respuesta: About 564 hours.
Ejemplo resuelto: The top 5%
What lifetime separates the top 5% of batteries from the rest? (The z with 95% to its left is $1.645$.)
- Top 5% above means 95% below: left area $0.95$, so $z = 1.645$.
- $x = 500 + 1.645(50) = 582.25$.
Respuesta: About 582 hours.
Ejemplo resuelto: The middle 80%
Which two lifetimes mark off the middle 80%? (The z with 10% to its left is $-1.28$.)
- The middle 80% leaves 10% in each tail.
- Lower cutoff: $z = -1.28$, so $x = 500 - 64 = 436$.
- Upper cutoff: by symmetry $z = 1.28$, so $x = 564$.
Respuesta: 436 and 564 hours.
Error común
Looking up the area of the tail you care about instead of the area to the left. For the top 5%, looking up $0.05$ gives $z = -1.645$, a value far below the mean. Check: a "top" cutoff must be above the mean, so its z must be positive.