**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
σ

^{2}= 2.25cm
σ = √2.25 = 1.5
cm

Let

*X*represents the diameter of an orange.*X*~

*N*(3.2, 1.5

^{2})

**(a)**

**(b)**

$\begin{array}{l}P\left(X<k\right)=0.305\\ P\left(Z<\frac{k-3.2}{1.5}\right)=0.305\\ \text{Fromthestandardnormaldistributiontable,}\\ P\left(Z0.51\right)=0.305\\ P\left(Z-0.51\right)=0.305\\ \frac{k-3.2}{1.5}=-0.51\\ k-3.2=-0.765\\ k=2.435\end{array}$

**Question 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)**

$\begin{array}{l}P\left(114<X<150\right)\\ =P\left(\frac{114-130}{16}<Z<\frac{150-130}{16}\right)\\ =P\left(-1<Z<1.25\right)\\ =1-P\left(Z>1\right)-P\left(Z>1.25\right)\\ =1-0.1587-0.1056\\ =0.7357\end{array}$

**(b)**

Probability
of getting grade ‘A’ tomatoes,

*P*(

*X*> 150) =

*P*(

*Z*> 1.25)

= 0.1056

$\begin{array}{l}\text{Letstotalnumberoftomatoes}=N\\ 0.1056\times N=132\\ N=\frac{132}{0.1056}\\ N=1250\end{array}$