Reading areas under the curve directly, using the normal cumulative distribution.
8 min read2 worked examples
The normal cumulative distribution
You can find probabilities for a normal distribution using the normal cumulative distribution function on your calculator. You enter the mean $\mu$, the standard deviation $\sigma$, and a lower and upper bound, and the function returns the area under the curve between those bounds.
Watch out
You need to enter a lower limit even when you only want an upper tail, and an upper limit even when you only want a lower tail. Choose a value at least $5$ standard deviations away from the mean — for example $\mu-100\sigma$ or $\mu+100\sigma$ — since the probability beyond that point is negligibly close to $0$.
Worked example 1
Four probabilities from one distribution
$X\sim \mathrm{N}(45, 6^2)$. Find: a $\mathrm{P}(X<52)$ b $\mathrm{P}(X\ge 36)$ c $\mathrm{P}(50.5<X<57)$ d $\mathrm{P}(X<40 \text{ or } X>50)$
a
$\mathrm{P}(X<52)=0.8783$ (4 d.p.). $52$ is larger than the mean, so the probability should be greater than $0.5$.
$\mathrm{P}(X<52) = 0.8783$ (4 d.p.).
b
$\mathrm{P}(X\ge 36)=\mathrm{P}(X>36)=0.9332$ (4 d.p.), since $\mathrm{P}(X=36)=0$ for a continuous distribution.
$\mathrm{P}(X\ge 36) = \mathrm{P}(X>36) = 0.9332$ (4 d.p.). You can use $>$ and $\ge$ interchangeably with a continuous distribution.
c
$\mathrm{P}(50.5<X<57)=0.1569$ (4 d.p.). Both bounds are above the mean, so the probability should be less than $0.5$.
$\mathrm{P}(50.5<X<57) = 0.1569$ (4 d.p.). Enter both bounds directly into the calculator.
d
$\mathrm{P}(40<X<50)=0.5953$, so $\mathrm{P}(X<40\text{ or }X>50)=1-0.5953=0.4047$ (4 d.p.).
1
$\mathrm{P}(X<40\text{ or }X>50) = 1-\mathrm{P}(40<X<50)$.
The total probability is $1$, so the two "tails" are whatever's left after removing the middle.
2
$\mathrm{P}(40<X<50) = 0.5953$, so $\mathrm{P}(X<40\text{ or }X>50) = 1-0.5953 = 0.4047$ (4 d.p.).
Sketch the two tails to check the answer looks sensible before committing to it.
▪ solved
Worked example 2
A normal model with a binomial extension · Challenge
The lengths, $L$ cm, of trout caught in a particular lake are modelled as $L\sim \mathrm{N}(35, 6^2)$. A trout longer than $50\,$cm is classified as a "trophy" catch. a Find the probability that a randomly caught trout is a trophy catch, to 3 s.f. b 15 trout are caught. Find the probability that at least 2 are trophy catches.
a
$\mathrm{P}(L>50)=0.00621$ (3 s.f.) — a small tail area, since $50$cm is $2.5$ standard deviations above the mean.
$\mathrm{P}(L>50) = 0.00621$ (3 s.f.).
b
MethodModel the count of trophy catches as binomial
1
Let $Y$ be the number of trophy catches among the 15 trout. $Y\sim \mathrm{B}(15, 0.00621)$.
Each of the 15 independent catches has the same probability of being a trophy catch.