Question

Nielsen Rating: The television show nbc Sunday night football broadcast a game between the colts and patriots and received a share of 22, meaning that among the tv sets in use, 22% are turned to that game (based on data from Nielsen Media Research). An advertiser wants to obtain a second opinion by conducting its own surveys and a pilot survey begins with 20 households having tv sets in use at the time of that same nbc Sunday night football broadcast.
a. Find the probability that none of the households are turned to nbc Sunday night football (3 digits)
b. Find the probability that at least one of the households is turned to nbc Sunday night football.(3 digits)
c. Find the probability that at most one of the households is turned to nbc Sunday night football (3 digits)
d. If at most one household is turned to nbc Sunday night football, does it appear that the 22% share value is wrong? Why or why not?

Answer 1

Answer:

a. 0.007 = 0.7% probability that none of the households are turned to nbc Sunday night football

b. 0.993 = 99.3% probability that at least one of the households is turned to nbc Sunday night football.

c. 0.046 = 4.6% probability that at most one of the households is turned to nbc Sunday night football

d. 1 is less than 2.5 standard deviations below the mean, so having at most one people watching shouldn't be considered an unusual outcome, which means that the 22% share values does not appear wrong.

Step-by-step explanation:

For each household, there are only two possible outcomes. Either it is tuned to Sunday Night Football, or it is not. The probability of a household being tuned to Sunday Night Football is independent of any other household. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

22% are turned to that game

This means that $p = 0.22$

Sample of 20 households:

This means that $n = 20$

a. Find the probability that none of the households are turned to nbc Sunday night football:

This is P(X = 0). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.22)^{0}.(0.78)^{20} = 0.007$

0.007 = 0.7% probability that none of the households are turned to nbc Sunday night football.

b. Find the probability that at least one of the households is turned to nbc Sunday night football.

This is:

$P(X \geq 1) = 1 - P(X = 0)$

Since P(X = 0) = 0.007

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.007 = 0.993$

0.993 = 99.3% probability that at least one of the households is turned to nbc Sunday night football.

c. Find the probability that at most one of the households is turned to nbc Sunday night football (3 digits)

This is

$P(X \leq 1) = P(X = 0) + P(X = 1)$. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.22)^{0}.(0.78)^{20} = 0.007$

$P(X = 1) = C_{20,1}.(0.22)^{1}.(0.78)^{19} = 0.039$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.007 + 0.039 = 0.046$

0.046 = 4.6% probability that at most one of the households is turned to nbc Sunday night football.

d. If at most one household is turned to nbc Sunday night football, does it appear that the 22% share value is wrong? Why or why not?

A measure is considered unusually low if it is more than two and a half standard deviations below the mean.

The mean is:

$E(X) = np = 20*0.22 = 4.4$

The standard deviation is:

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.22*0.78} = 1.8526$

4.4 - 2.5*1.8526 = -0.2.

1 is less than 2.5 standard deviations below the mean, so having at most one people watching shouldn't be considered an unusual outcome, which means that the 22% share values does not appear wrong.