Question

The television show CSI: Shoboygan 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 CSI: Shoboygan. Assume that an advertiser wants to verify that 23% share value by conducting its own survey, and a pilot survey begins with 15 households have TV sets in use at the time of a CSI: Shoboygan broadcast.

Find the probability that none of the households are tuned to CSI: Shoboygan.
P(none) =

Find the probability that at least one household is tuned to CSI: Shoboygan.
P(at least one) =

Find the probability that at most one household is tuned to CSI: Shoboygan.
P(at most one) =

If at most one household is tuned to CSI: Shoboygan, does it appear that the 23% share value is wrong? (Hint: Is the occurrence of at most one household tuned to CSI: Shoboygan unusual?)

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

Answer 1

Here, n = 15, p = 0.23, (1 - p) = 0.77 and x = 0
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 0)
P(X = 0) = 15C0 * 0.23^0 * 0.77^15
P(X = 0) = 0.0198

b)

Here, n = 15, p = 0.23, (1 - p) = 0.77 and x = 0
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 0).
P(X <= 0) = (15C0 * 0.23^0 * 0.77^15)
P(X <= 0) = 0.0198
P(X <= 0) = 0.0198

P(x> =1) = 1- P(x< =0)
= 1 - 0.0198
= 0.9802

c)

Here, n = 15, p = 0.23, (1 - p) = 0.77 and x = 1
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 1).
P(X <= 1) = (15C0 * 0.23^0 * 0.77^15) + (15C1 * 0.23^1 * 0.77^14)
P(X <= 1) = 0.0198 + 0.0889
P(X <= 1) = 0.1087

d)

no, it is not wrong