1. ## normal distribution

im having problem with this question:

The mass of a grade A papaya follows a normal distribution with mean 1.8 kg and standard deviation 200 g. The mass of a grade B papaya follows a normal distribution with mean 1.5 kg and standard deviation 140 g. Determine the probability that
a)3 grade A papayas chosen at random has a total mass of less than 5 kg,

b)3 times the mass of a grade A papaya chosen at random is less than 5 kg,

c)the total mass of 6 grade A papayas is more than that of 7 grade B papayas, all chosen at random.

2. This seems more like a prob or stats question than calculus but whatever. Let's call A the mass distribution of grade A papayas and B the mass distribution of grade B papayas. Well then,

$A\sim N(1.8,0.2^2)$ and $B\sim N(1.5, 0.14^2)$

a) We want 3 grade A papayas so the distribution is $A+A+A\sim N(1.8+1.8+1.8,0.2^2+0.2^2+0.2^2)=N(5.4,0.12)$. So, $\frac{A+A+A-5.4}{\sqrt{0.12}}\sim N(0,1)$. To find the probability, use a normal distribution table to find:

$
P(A+A+A<5)=\Phi (\frac{5-5.4}{\sqrt{0.12}})=\Phi (-1.15)=0.1251
$

b) Same logic, $3A\sim N(3(1.8),(3\cdot 0.2)^2)=N(5.4,0.36)$

$
P(3A<5)=\Phi (\frac{5-5.4}{\sqrt{0.36}})=\Phi (-0.67)=0.2514
$

c) Let X= A+A+...+A (six times) -B-B-...-B (seven times)

then, $X\sim N(6(1.8)-7(1.5), 6(0.2^2)+7(0.14^2))=N(0.3, 0.3772)$

So to have 6 A papayas heavier than 7 B papayas means that X>0.

$
P(X>0)=1-P(X<0)=1-\Phi (\frac{0-0.3}{\sqrt{0.3772}})=1-\Phi (-0.49)=1-0.3121=0.6879
$