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30.07.2019 •
Mathematics
Abinomial distribution has 100 trials (n = 100) with a probability of success of 0.25 (π = 0.25). to apply the normal distribution to approximate the binomial, what are the mean and standard deviation?
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Ответ:
E(X) = np = 100×0.25 = 25
probability of failure = 1-0.25=0.75
Variance[V(X)] = npq = 100×0.25×0.75 = 18.75
Standard deviation = √V(X) = √18.75 = 4.33
Ответ:
The fraction for 0.455555... is 41/90.
Step-by-step explanation:
Hope this helps!