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This is a intro to statistics question, I was wondering if i can have the work shown soni can see the process, thanks :)Charges for advertising on a TV show are based on the number of viewers, which is measured by the rating. The rating is a percentage of the population of 110 million TV households. The CBS television show 60 Minutes recently had a rating of 7.8, indicating that 7.8% of the households were tuned to that show. An advertiser conducts an independent survey of 100 households and finds that only 10 were tuned to 60 Minutes. Assuming that the 7.8 rating is correct, find the probability of surveying 100 randomly selected households and getting 10 or fewer tuned to 60 Minutes. What does this suggest? Does the advertiser have grounds for claiming a refund on the basis that the size of the audience was exaggerated?

User Ohadpr
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1 Answer

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Explanation:

7.8% were viewing the show.

so, when picking a single household the probability that is is watching the show is 0.078.

that also means that the probability that this household is not watching the show is 1 - 0.078 = 0.922.

"only" 10 of 100 were watching the show ? that is 10% and therefore a higher rate than measured by the rating agency.

there is no indication that the audience was exaggerated.

anyway, the probability to get 10 or fewer out of randomly picked 100 households to watch the show is

the probability of exactly 10 households watching +

the probability of exactly 9 households watching +

the probability of exactly 8 households watching +

...

the probability of exactly 1 household watching +

the probability of no (0) household watching.

to get the probabilty of any of these events :

x = number of households watching out of the overall 100.

w = probability of a household to watch = 0.078

a = probability of a household to be absent (= not watching) = 0.922

probability(x) = p(x) = w^x × a^(100-x) × c(100, x)

in other words :

the combined probability of "x" households watching and "a" households not watching (which must be 100-x) multiplied by the number of combinations of picking x households out of 100.

so, for the first one, x = 10, we get

p(10) = 0.078¹⁰ × 0.922⁹⁰ × 100! / (10! × (100 - 10)!) =

= 0.078¹⁰×0.922⁹⁰×100×99×...×91 / 10×9×8×...×2 =

= 8.335775831e-012 ×

0.00066955... ×

1.731030945644e+13 =

= 0.096612596...

the rest is then best via Excel or other spreadsheet applications.

p(9) = 0.125496...

p(8) = 0.145117...

p(7) = 0.147558...

p(6) = 0.129888...

p(5) = 0.096969...

p(4) = 0.059699...

p(3) = 0.0291

p(2) = 0.01053

p(1) = 0.002515...

p(0) = 0.000297...

and the sum of all that is the probability to have 10 or fewer households in randomly picked 100 watching the show :

0.843782...

the vast majority of samples of 100 households is expected to have 10 or fewer households watching the show.

this covers the general rating (that would suggest that 7.8 households of 100 are watching the show), some even better ranges (more than 7.8 households), but also everything that is worse than the original rating (below 7.8 households).

when we look at the individual probabilities, we see that the largest probabilities by far are in the categories of 9, 8, 7 and 6 households out of 100 are watching.

so, these are the main expected results when picking samples of 100 households.

therefore, I don't see any indication that the advertiser was "cheated".

User Henrik Joreteg
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