Question

Based on a Comcast survey, there's a 0.9 probability that a randomly selected adult or watch prime-time TV live, instead of online or on DVR. 16 adults are randomly selected.

(a). Find The probability that fewer than eight of the selected adults watch prime-time live.

(b). Find the mean number of adults watch prime-time live in samples of 16 adults. Do not round.

(c). find the standard deviation of adults who watch prime-time live in samples of 16 adults. Round to two decimal positions.

(d). Would it be unusual statistically significant for all of the 16 randomly selected adults to be watching prime time TV live?

Answer 1

n=16

p=0.9

a) P(X≤8) = Σ16C x * 0.9^ x * (1-0.9)^(16-x) where x varies from 0 to 8 = 0.000061
b) Mean = np =    16*0.9=       14.40

c) Standard deviation = √(np(1-p)) =   √(16*0.9*(1-0.9))=       1.2000

d) P ( X =    16   ) = C(16,16) * 0.9^16 * (1-0.9)^0 =            0.1853   (answer)

NO,  it would not be unusual ,

necause probability > 0.05