Question

The television show Law & Disorder has been successful for many years. That show recently had a share of 23, meaning that among the TV sets in use, 23% were tuned to Law & Disorder. Assume that an advertiser wants to verify that 23% share value by conducting its own survey, and a pilot survey begins with 13 households have TV sets in use at the time of a Law & Disorder broadcast.

Find the probability that none of the households are tuned to Law & Disorder.
P(none) =

Find the probability that at least one household is tuned to Law & Disorder.
P(at least one) =

Find the probability that at most one household is tuned to Law & Disorder.
P(at most one) =

If at most one household is tuned to Law & Disorder, does it appear that the 23% share value is wrong? (Hint: Is the occurrence of at most one household tuned to Law & Disorderunusual?)

• yes, it is wrong
• no, it is not wrong

Answer 1

Solution:

We are given that: The television show Law & Disorder has been successful for many years and among the TV sets in use, 23% were tuned to Law & Disorder

That is: p = 0.23

Sample size = n = 13

X= number of households are tuned to Law & Disorder follows Binomial distribution with n = 13 and p= 0.23

Part a) Find the probability that none of the households are tuned to Law & Disorder.

That is: P( None ) = P(X = 0) =.......?

Using Binomial probability formula:

where

and q = 1 - p = 1 - 0.23 = 0.77

Thus

Part b) Find the probability that at least one household is tuned to Law & Disorder.

From part a) P(X = 0) = 0.03345

Part c) Find the probability that at most one household is tuned to Law & Disorder.

From part a) P(X = 0) =0.03345

Now find P( X = 1)

Thus

Part d) If at most one household is tuned to Law & Disorder, does it appear that the 23% share value is wrong?

Since , is greater than 0.05, hence 23% share value is NOT wrong. Thus correct answer is:

No, it is not wrong.