# 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 220g and a variance of 100g. Find the probability that a pear that is picked at random has a mass
(a) of more than 230g.
(b) between 210g and 225g.
Hence, find the value of h such that 90% of the pears weigh more than h g.

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

(b)

For 90% (probability = 0.9) of the pears weigh more than h g,
P(X > h) = 0.9
P(X < h) = 1 – 0.9
= 0.1
From the standard normal distribution table,
P(Z > 0.4602) = 0.1
P(Z < –0.4602) = 0.1
$\begin{array}{l}\frac{h-220}{10}=-0.4602\\ h-220=-4.602\\ h=215.4\end{array}$