Normal distribution is fine approximation to the binomial distribution, in a binomial distribution one can easily confirm that the mean for a single binomial trial, where "success" is scored as 1 and "failure" is scored as 0, is p; where p is the probability of S. Where S is a sample space hence the mean for the binomial distribution with n trials is np. Condition to the normal approximation is good for the binomial distribution. Condition of failure for normal approximation is p (1-p), standard deviation for normal approximation (np(1-p))^{5}.

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If the number of trials in a sample space, the n is large, the binomial distribution is just about equal to the normal distribution. This is fine, since we actually do not feel like to plainly calculate binomial probabilities when n > 100.

Here, X is N( 0.7, 0.0002)

a) P(x > 0.71) = p(z > ($\frac{(0.71-0.7)}{0.01}$))

= p(z > 0.5) = 1 – p(0 < z < 1)

= 1 – 0.381 = 0.62

The result is 0.62

b) P[(x > 0.825) U (x < 0.785)] = 2P(x > 0.825)

= 2P(z > $\frac{0.025}{0.02}$)

= 2P(z > 1.25)

= 2[ -0.3944 + 0.5 ]

= 0.2112

Formula for cumulative binomial distribution

P(X = r) = nC

^{}

Given r = 3, n = 45, p = 0.08

Now we can substitute this values for this formula

= 45C_{3} *(.08)^{3}* (.92)^{45-3}

= 45C_{3} *(.08)^{3}* (.92)^{42}

= 0.219

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