Probability Distributions, Long Questions (Question 4 & 5)


Question 4:
The diameter of oranges harvested from a fruit orchard has a normal distribution with a mean of 3.2 cm and a variance of 2.25 cm.
Calculate
(a) the probability that an orange chosen at random from this fruit orchard has a diameter of more than 3.8 cm.
(b) the value of k if 30.5 % of the oranges have diameter less than k cm.

Solution:
µ = 3.2 cm
σ2 = 2.25cm
σ = √2.25 = 1.5 cm
Let X represents the diameter of an orange.
X ~ N (3.2, 1.52)
(a)
P( X>3.8 ) =P( Z> 3.83.2 1.5 ) =P( Z>0.4 ) =0.3446

(b)
P( X<k )=0.305 P( Z< k3.2 1.5 )=0.305 From the standard normal distribution table, P( Z>0.51 )=0.305 P( Z<0.51 )=0.305 k3.2 1.5 =0.51 k3.2=0.765 k=2.435


Question 5:
The masses of tomatoes in a farm are normally distributed with a mean of 130g and standard deviation of 16g. Tomato with weight more than 150g is classified as grade ‘A’.
(a) A tomato is chosen at random from the farm.
Find the probability that the tomato has a weight between 114g and 150g. 
(b) It is found that 132 tomatoes in this farm are grade ‘A’.
Find the total number of tomatoes in the farm.

Solution:
µ = 130
σ = 16
(a)
P( 114<X<150 ) =P( 114130 16 <Z< 150130 16 ) =P( 1<Z<1.25 ) =1P( Z>1 )P( Z>1.25 ) =10.15870.1056 =0.7357
(b)
Probability of getting grade ‘A’ tomatoes,
P(X > 150) = P(Z > 1.25)
                    = 0.1056
Lets total number of tomatoes=N 0.1056×N=132 N= 132 0.1056 N=1250