Question 1103988
In order for the normal approximation to work, the following conditions must hold:
{{{n*p >= 5}}}
{{{n*(1-p) >= 5}}}
Note: Some books use 10 in place of 5


Compute both n*p and n*(1-p)
n*p = 5*0.3 = 1.5
n*(1-p) = 5*(1-0.3) = 5*0.7 = 3.5


So both {{{n*p >= 5}}} and {{{n*(1-p) >= 5}}} are false. The same applies to {{{n*p >= 10}}} and {{{n*(1-p) >= 10}}} as well.


Therefore the normal approximation cannot be used.