document.write( "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.
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Algebra.Com's Answer #812105 by mathmate(429)\"\" \"About 
You can put this solution on YOUR website!
Given:
\n" ); document.write( "a sample of 145 seeds whose probability of germination is about 0.74.
\n" ); document.write( "Owner wants to know the probability that at least 104 seeds would germinate.\r
\n" ); document.write( "\n" ); document.write( "Solution:
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\n" ); document.write( "Conditions of applying binomial distribution:
\n" ); document.write( "- Bernoulli trian (success or failure)
\n" ); document.write( "- Number of trials known (n=145)
\n" ); document.write( "- probability of success is known (p=0.74) and remains constant
\n" ); document.write( "- all trials are independent.
\n" ); document.write( "All satisfied, so will apply binomial distribution.
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\n" ); document.write( "Binomial distribution formula for calculating probability of success of x trials out of n, each with probability p:
\n" ); document.write( "P(x,n,p)= C(n,x)p^(x)p^(n-x)
\n" ); document.write( "where
\n" ); document.write( "C(n,x) = n!/(x!(n-x)!) = combination of selecting x items out of n
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\n" ); document.write( "Using
\n" ); document.write( "p = 0.74
\n" ); document.write( "n = 145
\n" ); document.write( "x = 104 to 145
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\n" ); document.write( "Taking advantage of today's ease with technology, we will be able to calculate and sum large number of cases without much difficulties.
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\n" ); document.write( "S = P(x,n,p) for x = 104 to 145, with the following results:
\n" ); document.write( " x cumulative P(x,n,p)
\n" ); document.write( "104 0.060599 0.0605989
\n" ); document.write( "105 0.127946 0.067347
\n" ); document.write( "106 0.200278 0.072332
\n" ); document.write( "107 0.275314 0.075036
\n" ); document.write( "108 0.350457 0.0751429
\n" ); document.write( "109 0.423054 0.0725975
\n" ); document.write( "110 0.490676 0.0676223
\n" ); document.write( "111 0.551363 0.0606866
\n" ); document.write( "112 0.603797 0.0524339
\n" ); document.write( "113 0.647379 0.0435819
\n" ); document.write( "114 0.682198 0.0348185
\n" ); document.write( "115 0.708911 0.0267135
\n" ); document.write( "116 0.728574 0.0196631
\n" ); document.write( "117 0.742446 0.0138715
\n" ); document.write( "118 0.751814 0.00936825
\n" ); document.write( "119 0.757864 0.0060497
\n" ); document.write( "120 0.761594 0.00373065
\n" ); document.write( "121 0.763788 0.0021938
\n" ); document.write( "122 0.765017 0.0012283
\n" ); document.write( "123 0.765670 6.53714*10^-4
\n" ); document.write( "124 0.766000 3.30101*10^-4
\n" ); document.write( "125 0.766158 1.57839*10^-4
\n" ); document.write( "126 0.766230 7.13071*10^-5
\n" ); document.write( "127 0.766260 3.03627*10^-5
\n" ); document.write( "128 0.766272 1.21524*10^-5
\n" ); document.write( "129 0.766277 4.55805*10^-6
\n" ); document.write( "130 0.766278 1.59666*10^-6
\n" ); document.write( "131 0.766279 5.20347*10^-7
\n" ); document.write( "132 0.766279 1.57074*10^-7
\n" ); document.write( "133 0.766279 4.36974*10^-8
\n" ); document.write( "134 0.766279 1.11375*10^-8
\n" ); document.write( "135 0.766279 2.5829*10^-9
\n" ); document.write( "136 0.766279 5.40539*10^-10
\n" ); document.write( "137 0.766279 1.01066*10^-10
\n" ); document.write( "138 0.766279 1.66754*10^-11
\n" ); document.write( "139 0.766279 2.39011*10^-12
\n" ); document.write( "140 0.766279 2.91541*10^-13
\n" ); document.write( "141 0.766279 2.94245*10^-14
\n" ); document.write( "142 0.766279 2.35906*10^-15
\n" ); document.write( "143 0.766279 1.40858*10^-16
\n" ); document.write( "144 0.766279 5.56812*10^-18
\n" ); document.write( "145 0.766279 1.09294*10^-19
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\n" ); document.write( "Therefore the probability of 104 and more seeds germinating as predicted by the binomial distribution is 0.76628.
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\n" ); document.write( "Normal Approximation
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\n" ); document.write( "Unfortunately p = 0.74 is, a little skew (from 0.5), although acceptable. The skew will result in a less exact approximation, continue reading!
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\n" ); document.write( "Parameters of the binomial distribution:
\n" ); document.write( "mean = np = 145*0.74 = 107.3
\n" ); document.write( "variance = npq = 145*0.74*(1-0.74) = 27.898
\n" ); document.write( "standard deviation = sqrt(variance) = 5.281856
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\n" ); document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "Calculate z(x=103.5)
\n" ); document.write( "z = (103.5-107.3)/5.281856 = -0.719444
\n" ); document.write( "Looking up the standard normal probability table provides
\n" ); document.write( "P(x>=103.5) = P(z>=-0.719444) = 1 - P(z<=-0.719444) = 1 - 0.2359337 = 0.7640663\r
\n" ); document.write( "\n" ); document.write( "Compared to the binomial distribution results (0.7662794), there is a noticeable error of -0.0022, or 0.29%. \n" ); document.write( "
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