8.4.2 Probability Distributions, Long Questions


Question 5:
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
σ= 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 > 3.8 3.2 1.5 ) = P ( Z > 0.4 ) = 0.3446

(b)
P ( X < k ) = 0.305 P ( Z < k 3.2 1.5 ) = 0.305 From the standard normal distribution table, P ( Z > 0.51 ) = 0.305 P ( Z < 0.51 ) = 0.305 k 3.2 1.5 = 0.51 k 3.2 = 0.765 k = 2.435



Question 6:
The masses of tomatoes in a farm are normally distributed with a mean of 130 g and standard deviation of 16 g. Tomato with weight more than 150 g 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 114 g and 150 g. 

(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 ( 114 130 16 < Z < 150 130 16 ) = P ( 1 < Z < 1.25 ) = 1 P ( Z > 1 ) P ( Z > 1.25 ) = 1 0.1587 0.1056 = 0.7357


(b)
Probability of getting grade ‘A’ tomatoes,
(X > 150) = (Z > 1.25)
= 0.1056
Lets total number of tomatoes = N 0.1056 × N = 132 N = 132 0.1056 N = 1250


Question 7:
In a boarding school entry exam, 300 students sat for a mathematics test. The marks obtained follow a normal distribution with a mean of 56 and a standard deviation of 8.

(a) Find the number of students who pass the test if the passing mark is 40.

(b) If 12% of the students pass the test with grade A, find the minimum mark to obtain grade A.

Solution:
Let X=marks obtained by students X~N( 56, 8 2 ) ( a ) P( X40 )=P( Z 4056 8 )    =P( Z2 )    =1P( Z2 )    =10.02275    =0.9773 Number of students who pass the test =0.9773×300 =293 ( b ) Let the minimum mark to obtain grade A be k P( Xk )=0.12 P( Z k56 8 )=0.12   k56 8 =1.17   k=( 1.17 )( 8 )+56 =65.36

Thus, the minimum mark to obtain grade A is 66.

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