What is StartFraction 7 Over 9 EndFraction divided by one-third?
2 answers:
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Answer: 0.951%
Explanation:
Note that in the problem, the scenario is either the adult is using or not using smartphones. So, we have a yes or no scenario involved with the random variable, which is the number of adults using smartphones. Thus, the number of adults using smartphones follows the binomial distribution.
Let x be the number of adults using smartphones and n be the number of randomly selected adults. In Binomial distribution, the probability that there are k adults using smartphones is given by
Where p = probability that an adult is using smartphones = 54% (since 54% of adults are using smartphones).
Since n = 12 and k = 3, the probability that fewer than 3 are using smartphones is given by
Therefore, the probability that there are fewer than 3 adults are using smartphone is 0.00951 or 0.951%.
Explanation:
Note that in the problem, the scenario is either the adult is using or not using smartphones. So, we have a yes or no scenario involved with the random variable, which is the number of adults using smartphones. Thus, the number of adults using smartphones follows the binomial distribution.
Let x be the number of adults using smartphones and n be the number of randomly selected adults. In Binomial distribution, the probability that there are k adults using smartphones is given by
Where p = probability that an adult is using smartphones = 54% (since 54% of adults are using smartphones).
Since n = 12 and k = 3, the probability that fewer than 3 are using smartphones is given by
Therefore, the probability that there are fewer than 3 adults are using smartphone is 0.00951 or 0.951%.