Let z = 19 + i and w = 4 + 10i.
2 answers:
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Using the binomial distribution, the probabilities are given as follows:
- 0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.
- 0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.
- Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.
The formula is:
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
The values of the parameters for this problem are:
n = 10, p = 0.4.
The probability that more than 4 weigh more than 20 pounds is:
In which:
Then:
Hence:
0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.
The probability that fewer than 3 weigh more than 20 pounds is:
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0061 + 0.0403 + 0.1209 = 0.1673
0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.
For more than 7, the probability is:
Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.
More can be learned about the binomial distribution at brainly.com/question/24863377
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Answer:
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