Answer to Question #86257 in Statistics and Probability for Anand

Question

Previous studies on some spherical seeds have revealed that their mean diameter is 10 mm with a standard deviation of 2 mm. We start with 1000 seeds and pass them through two sieves so that only seeds whose diameter is between 9.5mm and10.5mm are left. Find out the following:
(i) How many such seeds will we get?
(ii) If we discard only those seeds with diameter less than 6 mm, then how many will be left?

Answer 1

In[1]:= seeds = NormalDistribution[10, 2]
Out[1]= NormalDistribution[10, 2]

(i) Seeds within range from 9.5 mm to 10.5 mm diameter: 

In[6]:= 1000*Probability[9.5 < x < 10.5, x \[Distributed] seeds]
Out[6]= 197.413

(ii) Seeds larger than 6 mm diameter 

(i.e. within range from 6 mm to infinity): 

In[7]:= 1000*Probability[6 < x < Infinity, x \[Distributed] seeds]
Out[7]= 500 (1 + Erf[Sqrt[2]])
In[8]:= 1000*N[Probability[6 < x < Infinity, x \[Distributed] seeds]]
Out[8]= 977.25

For more knowledge on methodology, please refer to e.g.:

http://mathworld.wolfram.com/NormalDistribution.html