Question

In: Statistics and Probability

A recent Wall Street Journal article indicated that 37% of American teenagers use Pandora's online radio...

A recent Wall Street Journal article indicated that 37% of American teenagers use Pandora's online radio service. Suppose a sample of 25 American teenagers is selected. Based on this information, generate a cumulative binomial probability distribution in the given space. Round to four decimals for the answer.

Find the probability that the number of American teenagers who use Pandora's online radio service differs by greater than 1 from the mean.

Find the probability that greater than 4 American teenagers use Pandora's online radio service.

Find the probability that no less than 8 American teenagers use Pandora's online radio service.

Find the probability that more than 25% but at most 60% of these American teenagers use Pandora's online radio service.

Find the probability that at least 14 American teenagers do not use Pandora's online radio service.

Find the probability that no more than 5 American teenagers use Pandora's online radio service.

Expert Solution

here this is binomial with parameter n=25 and p=0.37

mean E(x)=μ=np=9.25

9.25

standard deviation σ=√(np(1-p))=2.414

1)

probability that the number of American teenagers who use Pandora's online radio service differs by greater than 1 from the mean =1-P(7<=X<=11) =1-0.6991 =0.3009

2)

probability that greater than 4 American teenagers use Pandora's online radio service

P(X>4)=1-P(X<=4)=

1-∑_x=0^x-1 (_nC_x)p^x(q)^(n-x) =

0.9799

4)

Find the probability that no less than 8 American teenagers use Pandora's online radio service.

P(X<8)=

∑_x=0⁷ (_nC_x)p^x(1−p)^(n-x) =

0.2374

4)

probability that more than 25% but at most 60% of these American teenagers use Pandora's online radio service:

P(7<=X<=15)=

∑_x=a^b (_nC_x)p^x(1−p)^(n-x) =

0.8686

0.1314

5)

probability that at least 14 American teenagers do not use Pandora's online radio service

P(X<=11)=

∑_x=0^a (_nC_x)p^x(1−p)^(n-x) =

0.8249

6)

P(X<=5)=

∑_x=0^a (_nC_x)p^x(1−p)^(n-x) =

0.0559

orchestra answered 2 years ago

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