The television show found has been successful for many years. that show recently had a share of 19, meaning that among the tv sets in use, 19% were tuned to found. assume that an advertiser wants to verify that 19% share value by conducting its own survey, and a pilot survey begins with 14 households have tv sets in use at the time of a found broadcast. find the probability that none of the households are tuned to found. p(none) =(round answer to 4 decimal places)find the probability that at least one household is tuned to found. p(at least one) =(round answer to 4 decimal places)find the probability that at most one household is tuned to found. p(at most one) =(round answer to 4 decimal places)if at most one household is tuned to found, does it appear that the 19% share value is wrong? (hint: is the occurrence of at most one household tuned to found unusual? )yes, it is wrongno, it is not wrong

a) 0.0523 = 5.23%

b) 0.9477 = 94.77%

c) 0.2242 = 22.42%

d) YES

Step-by-step explanation:

This situation could be modeled with a binomial distribution where

The probability of getting exactly k “successes” in n trials is given by  

Where p is the probability of “success”.

In this case we can assume that “success” is the fact that the household interviewed is tuned to Found.  

So, p=0.19 and n = 14 (the households interviewed).

a)

The probability that none of the households are tuned to Found is P(X=0)

b)

The probability that at least one household is tuned to Found is  

1- P(X=0) = 1-0.0523 = 0.9477 = 94.77%

c)

The probability that at most one household is tuned to Found is  P(X=0)+P(X=1)

d)

According to this sample, the probability that more than one household is tuned to Found would be 100%-22.42% = 77.58%, so it does appear that the 19% share value is wrong.

i think it would be 42cm

step-by-step explanation: