# 8.2c Probability of an Event

8.2c Probability of an Event

Example:
The masses of pears in a fruit stall are normally distributed with a mean of 220 g and a variance of 100 g. Find the probability that a pear that is picked at random has a mass
(a) of more than 230 g.
(b) between 210 g and 225 g.
Hence, find the value of h such that 90% of the pears weigh more than h g.

Solution:
µ = 220 g
σ = √100 = 10 g
Let X be the mass of a pear.

(a)

(b)

For 90% (probability = 0.9) of the pears weigh more than h g,
(X > h) = 0.9
(X < h) = 1 – 0.9
= 0.1

From the standard normal distribution table,
(Z > 0.4602) = 0.1
(Z < –0.4602) = 0.1
$\begin{array}{l}\frac{h-220}{10}=-0.4602\\ h-220=-4.602\\ h=215.4\end{array}$