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Advanced Math/The Normal approximation to binomial distribution

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Expert: - 6/13/2009


Question
in a series of n independent trials the probability of a 'success' at each trial is p. If R is the random variable denoting the total number of successes, state the probability that R=r. State the mean and variance of R. A certain variety of flower seed is sold in packets containing about 1000 seeds. The packet claims that 40% will bloom white & 60% red. This may be assumed to be accurate. If five seeds are planted estimate the probability that a) exactly 3 will bloom white b) at least 1 will bloom white. If 100 seeds are planted use the normal approximation to estimate the probability of obtaining between 30 & 45 white flowers. Thanks

Answer
For (a), the binomial can be used.  Calculate it for k = 3.

For (b), calculate 1 - P(all red).  
For this one, not that it is 0.6^5.

Use the normal with mean = np.  Here, n = 5 and p = 0.4.
The variance should be np(1-p).  A success is a white bloom.

If 100 seeds are planted, the average number of  seeds that are white can be approximated by the normal.
Our success rate would be 0.40 and the number would be 100.

To find the chances of a selection being between 30 and 45, find how many standard deviations 29.5 and 45.5 are from the mean, which is 40.  For each value, find the probability of being between 0 and that number in a Normal distribution, then add them together.  Note: If the normal table goes from –infinity to the number, subtract 0.5 from both values.  The answer for each should be between 0 and 0.5.

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