The length of a social media interaction is normally distributed with a mean of 3 minutes and a standard deviation of .4 minutes
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Answer is Choice C
I used the power rule for integrals.
I used the power rule for integrals.
Using the binomial distribution, it is found that there is a:
a) The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.
b) The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.
c) The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.
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For each chipmunk, there are only two possible outcomes. Either they will live to be 4 years old, or they will not. The probability of a chipmunk living is independent of any other chipmunk, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
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.
In this problem:
- 0.96516 probability of a chipmunk living through the year, thus
Item a:
- Two is P(X = 2) when n = 2, thus:
The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.
Item b:
- Six is P(X = 6) when n = 6, then:
The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.
Item c:
- At least one not living is:
The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.
A similar problem is given at brainly.com/question/24756209