Question 1182076: A researcher believes that about 74% of the seeds planted with the aid of a new chemical fertilizer will germinate. He chooses a random sample of 145 seeds and plants them with the aid of the fertilizer. Assuming his belief to be true, approximate the probability that at least 104 of the 145 seeds will germinate. Use the normal approximation to the binomial with a correction for continuity.
Round your answer to at least three decimal places. Do not round any intermediate steps.
Found 2 solutions by Boreal, mathmate:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! np=145*0.74=107.3=mean in pounds
np(1-p)=variance=107.3*0.26=27.898
sd=sqrt(V)=5.28 pounds
correction factor will be n-0.5 or 103.5
z=(103.5-107.3)/5.28
>=-3.8/4.28
=-0.8878
that probability is 0.8127
Answer by mathmate(429)  (Show Source):
You can put this solution on YOUR website! 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%.
|
|
|
