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41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U
Question
41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four.
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Mathematics
3 years
2021-09-05T13:53:05+00:00
2021-09-05T13:53:05+00:00 1 Answers
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Answers ( )
Answer:
a) 0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.
b) 0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.
c) 0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.
Step-by-step explanation:
For each adult, there are only two possible outcomes. Either they have very little confidence in newspapers, or they do not. The answers of each adult are independent, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
41% of U.S. adults have very little confidence in newspapers.
This means that![Rendered by QuickLaTeX.com p = 0.41](https://documen.tv/wp-content/ql-cache/quicklatex.com-a76bd69c36c0f4cc82e00494713bdef0_l3.png)
You randomly select 10 U.S. adults.
This means that![Rendered by QuickLaTeX.com n = 10](https://documen.tv/wp-content/ql-cache/quicklatex.com-a545146ef1b6216866958e2b6d2fc003_l3.png)
(a) exactly five
This is
. So
0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.
(b) at least six
This is:
Then
0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.
(c) less than four.
This is:
So
0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.