Question 1182076
Given:
a sample of 145 seeds whose probability of germination is about 0.74.
Owner wants to know the probability that at least 104 seeds would germinate.

Solution:
 
Conditions of applying binomial distribution:
- Bernoulli trian (success or failure)
- Number of trials known (n=145)
- probability of success is known (p=0.74) and remains constant
- all trials are independent.
All satisfied, so will apply binomial distribution.
 
Binomial distribution formula for calculating probability of success of x trials out of n, each with probability p:
P(x,n,p)= C(n,x)p^(x)p^(n-x)
where
C(n,x) = n!/(x!(n-x)!) = combination of selecting x items out of n
 
Using
p = 0.74
n = 145
x = 104 to 145
 
Taking advantage of today's ease with technology, we will be able to calculate and sum large number of cases without much difficulties.
 
S = P(x,n,p) for x = 104 to 145, with the following results:
 x  cumulative P(x,n,p)
104  0.060599  0.0605989  
105  0.127946  0.067347  
106  0.200278  0.072332  
107  0.275314  0.075036  
108  0.350457  0.0751429  
109  0.423054  0.0725975  
110  0.490676  0.0676223  
111  0.551363  0.0606866  
112  0.603797  0.0524339  
113  0.647379  0.0435819  
114  0.682198  0.0348185  
115  0.708911  0.0267135  
116  0.728574  0.0196631  
117  0.742446  0.0138715  
118  0.751814  0.00936825  
119  0.757864  0.0060497  
120  0.761594  0.00373065  
121  0.763788  0.0021938  
122  0.765017  0.0012283  
123  0.765670  6.53714*10^-4  
124  0.766000  3.30101*10^-4  
125  0.766158  1.57839*10^-4  
126  0.766230  7.13071*10^-5  
127  0.766260  3.03627*10^-5  
128  0.766272  1.21524*10^-5  
129  0.766277  4.55805*10^-6  
130  0.766278  1.59666*10^-6  
131  0.766279  5.20347*10^-7  
132  0.766279  1.57074*10^-7  
133  0.766279  4.36974*10^-8  
134  0.766279  1.11375*10^-8  
135  0.766279  2.5829*10^-9  
136  0.766279  5.40539*10^-10  
137  0.766279  1.01066*10^-10  
138  0.766279  1.66754*10^-11  
139  0.766279  2.39011*10^-12  
140  0.766279  2.91541*10^-13  
141  0.766279  2.94245*10^-14  
142  0.766279  2.35906*10^-15  
143  0.766279  1.40858*10^-16  
144  0.766279  5.56812*10^-18  
145  0.766279  1.09294*10^-19  
 
Therefore the probability of 104 and more seeds germinating as predicted by the binomial distribution is 0.76628.
 
 
 
Normal Approximation
====================
 
Unfortunately p = 0.74 is, a little skew (from 0.5), although acceptable.  The skew will result in a less exact approximation, continue reading!
 
Parameters of the binomial distribution:
mean = np = 145*0.74 = 107.3
variance = npq = 145*0.74*(1-0.74) = 27.898
standard deviation = sqrt(variance) = 5.281856
 
Since binomial distribution is discrete, the continuity correction needs to be applied to improve accuracy of the approximation.  This can be done by calculating the probability of x between 103 and 104, namely 103.5, and 145, which maps to infinity in the normal distribution.

Calculate z(x=103.5)
z = (103.5-107.3)/5.281856 = -0.719444
Looking up the standard normal probability table provides
P(x>=103.5) = P(z>=-0.719444) = 1 - P(z<=-0.719444) = 1 - 0.2359337 = 0.7640663

Compared to the binomial distribution results (0.7662794), there is a noticeable error of  -0.0022, or 0.29%.