Respuesta :
Answer:
(a) The value of P (None) is 0.062.
(b) The value of P(at least one) is 0.938.
(c) The value of P(at most one) is 0.253.
(d) The event is not unusual.
Step-by-step explanation:
Let X = number of households watching the show.
The probability of the random variable x is, P (X) = p = 0.18.
The sample selected for the survey is of size, n = 14
The random variable X follows a Binomial distribution with parameter n = 14 and p = 0.18.
The probability of a Binomial distribution is computed using the formula:
[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,...[/tex]
(a)
Compute the probability that none of the households are tuned to CSI: Shoboygan as follows:
[tex]P(X=0)={14\choose 0}(0.18)^{0}(1-0.18)^{14-0}=1\times1\times0.06214=0.062[/tex]
Thus, the value of P (None) is 0.062.
(b)
Compute the probability that at least one household is tuned to CSI: Shoboygan as follows:
P (X ≥ 1) = 1 - P (X < 1)
= 1 - P (X = 0)
[tex]=1-0.062\\=0.938[/tex]
Thus, the value of P(at least one) is 0.938.
(c)
Compute the probability that at most one household is tuned to CSI: Shoboygan as follows:
P (X ≤ 1) = P (X = 0) + P (X = 1)
[tex]={14\choose 0}(0.18)^{0}(1-0.18)^{14-0}+{14\choose 1}(0.18)^{1}(1-0.18)^{14-1}\\=0.062+0.191\\=0.253[/tex]
Thus, the value of P(at most one) is 0.253.
(d)
An event that has a very low probability of occurrence is known as an unusual event.
The probability of the event "at most one household is tuned to CSI: Shoboygan" is 0.253.
This probability value is not low.
Hence, the event is not unusual.
