The weight limit, w, on the bridge is 12 tons.
1 answer:
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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 <em>X</em> = number of consumer's who recognize the brand.
The probability of the random variable <em>X</em> is, P (X) = <em>p</em> = 0.40.
A random sample of size, <em>n</em> = 6 consumers are selected.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.
The probability mass function of <em>X</em> is:
(a)
Compute the value of P (X = 5) as follows:
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:
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.