Respuesta :
Answer:
(a) The probability that exactly 5 of the 6 consumers recognize the brand name is 0.0369.
(b) The probability that all of the selected consumers recognize the brand name is 0.0041.
(c) The probability that at least 5 of the selected consumers recognize the brand name is 0.041.
(d) The events of 5 customers recognizing the brand name is unusual.
Step-by-step explanation:
Let X = number of consumer's who recognize the brand.
The probability of the random variable X is, P (X) = p = 0.40.
A random sample of size, n = 6 consumers are selected.
The random variable X follows a Binomial distribution with parameters n and p.
The probability mass function of X is:
[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...[/tex]
(a)
Compute the value of P (X = 5) as follows:
[tex]P(X=5)={6\choose 5}0.40^{5}(1-0.40)^{6-5}\\=6\times0.01024\times0.60\\=0.036864\\\approx0.0369[/tex]
Thus, the probability that exactly 5 of the 6 consumers recognize the brand name is 0.0369.
(b)
Compute the value of P (X = 6) as follows:
[tex]P(X=6)={6\choose 6}0.40^{6}(1-0.40)^{6-6}\\=1\times 0.004096\times1\\=0.004096\\\approx0.0041[/tex]
Thus, the probability that all of the selected consumers recognize the brand name is 0.0041.
(c)
Compute the value of P (X ≥ 5) as follows:
P (X ≥ 5) = P (X = 5) + P (X = 6)
= 0.0369 + 0.0041
= 0.041
Thus, the probability that at least 5 of the selected consumers recognize the brand name is 0.041.
(d)
An event is considered unusual if the probability of its occurrence is less than 0.05.
The probability of 5 customers recognizing the brand name is 0.0369.
This probability value is less than 0.05.
Thus, the events of 5 customers recognizing the brand name is unusual.
