**8.2c Probability of an Event**

Example:

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:Solution:

µ

*=*220 gσ = √100 = 10 g

Let

*X*be the mass of a pear.

(a)

(a)

$\begin{array}{l}P\left(X>230\right)\\ =P\left(Z>\frac{230-220}{10}\right)\leftarrow \overline{)\begin{array}{l}\text{Converttostandardnormal}\\ \text{distributionusing}Z=\frac{X-\mu}{\sigma}\end{array}}\\ =P\left(Z1\right)\\ =0.1587\end{array}$

(b)

(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

$\begin{array}{l}\frac{h-220}{10}=-0.4602\\ h-220=-4.602\\ h=215.4\end{array}$