Question 1084249
np=5, expected value
np(1-p)=5*0.95=4.75=variance
sd is sqrt(4.75)=2.18
The problem asks for the approximate probability.
an approximation using the continuity correction factor would be
between 5 and 10.5
z=(10.5-5)/2.18=+2.52
z for 5=0
want probability that z is not greater than +2.52
That is 0.0059
The approximate probability asked for is 0.9941