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)
P(X>230)=P(Z>23022010)Convert to standard normal distribution using Z=Xμσ=P(Z>1)=0.1587


(b)
P(210<X<225)=P(21022010<Z<22522010)Convert to standard normal distribution using Z=Xμσ=P(1<Z<0.5)=1P(Z>1)P(Z>0.5)=10.15870.3085=0.5328

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
h22010=0.4602h220=4.602h=215.4