A binomial experiment consists of 500 trials. The probability of success for each trial is .4 . What is the probability of obtaining the number of successes indicated in Problems 51-58? Approximate these probabilities to two decimal places using a normal curve. (This binomial experiment easily passes the rule-of-thumb test, as you can check. When computing the probabilities, adjust the intervals as in Examples 3 and 4.) 188 or more
A binomial experiment consists of 500 trials. The probability of success for each trial is .4 . What is the probability of obtaining the number of successes indicated in Problems 51-58? Approximate these probabilities to two decimal places using a normal curve. (This binomial experiment easily passes the rule-of-thumb test, as you can check. When computing the probabilities, adjust the intervals as in Examples 3 and 4.) 188 or more
Solution Summary: The author calculates the probability of obtaining 188 or less successes in a binomial experiment consisting of 500 trials.
A binomial experiment consists of
500
trials. The probability of success for each trial is
.4
. What is the probability of obtaining the number of successes indicated in Problems 51-58? Approximate these probabilities to two decimal places using a normal curve. (This binomial experiment easily passes the rule-of-thumb test, as you can check. When computing the probabilities, adjust the intervals as in Examples 3 and 4.)
Could you please help me with question 2bii. If possible could you explain how you found the bounds of the integral by using a graph of the region of integration. Thanks
Let A be a vector space with basis 1, a, b. Which (if any) of the following rules
turn A into an algebra? (You may assume that 1 is a unit.)
(i) a² = a, b² = ab = ba = 0.
(ii) a²=b, b² = ab = ba = 0.
(iii) a²=b, b² = b, ab = ba = 0.
Elementary Statistics: Picturing the World (7th Edition)
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Discrete Distributions: Binomial, Poisson and Hypergeometric | Statistics for Data Science; Author: Dr. Bharatendra Rai;https://www.youtube.com/watch?v=lHhyy4JMigg;License: Standard Youtube License