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5.2c Probability of an Event


5.2c 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> 230−220 10 )← Convert to standard normal  distribution using Z= X−μ σ =P( Z>1 ) =0.1587



(b)
P( 210<X<225 ) =P( 210−220 10 <Z< 225−220 10 )← Convert to standard normal  distribution using Z= X−μ σ =P( −1<Z<0.5 ) =1−P( Z>1 )−P( Z>0.5 ) =1−0.1587−0.3085 =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
h−220 10 =−0.4602 h−220=−4.602 h=215.4



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