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)
P( X>230 ) =P( Z> 230220 10 ) Convert to standard normal  distribution using Z= Xμ σ =P( Z>1 ) =0.1587  

(b)
P( 210<X<225 ) =P( 210220 10 <Z< 225220 10 ) Convert to standard normal  distribution using Z= Xμ σ =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, 
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
h220 10 =0.4602 h220=4.602 h=215.4