5.4.2 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)
$\begin{array}{l}P\left(X>3.8\right)\\ =P\left(Z>\frac{3.8-3.2}{1.5}\right)\\ =P\left(Z>0.4\right)\\ =0.3446\end{array}$

(b)
$\begin{array}{l}P\left(X0.51\right)=0.305\\ P\left(Z<-0.51\right)=0.305\\ \frac{k-3.2}{1.5}=-0.51\\ k-3.2=-0.765\\ k=2.435\end{array}$

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)
$\begin{array}{l}P\left(1141\right)-P\left(Z>1.25\right)\\ =1-0.1587-0.1056\\ =0.7357\end{array}$

(b)
Probability of getting grade ‘A’ tomatoes,
(X > 150) = (Z > 1.25)
= 0.1056
$\begin{array}{l}\text{Lets total number of tomatoes}=N\\ 0.1056×N=132\\ N=\frac{132}{0.1056}\\ N=1250\end{array}$

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:

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