Respuesta :
Answer:
Explained below.
Step-by-step explanation:
Let X = number of trucks that will be going over the speed limit on I-77 South between Dobson and Elkin.
The probability of X is, p = 0.75.
A random sample of n = 5 trucks are observed on this stretch of I-77.
The random variable X follows a binomial distribution with parameters n = 5 and p = 0.75.
(a)
Compute the mean as follows:
[tex]\mu=n\times p=5\times 0.75=3.75[/tex]
The mean of this probability distribution is 3.75.
(b)
The mean of 3.75 implies that on average 3.75 trucks that will be going over the speed limit on I-77 South between Dobson and Elkin.
(c)
Compute the standard deviation as follows:
[tex]\sigma=\sqrt{np(1-p)}=\sqrt{5\times 0.75\times (1-0.75)}=0.968[/tex]
Thus, the standard deviation of this probability distribution is 0.97.
(d)
Compute the probability that exactly 3 of the observed trucks are speeding as follows:
[tex]P(X=3)={5\choose 3}(0.75)^{3}(1-0.75)^{5-3}\\\\=10\times 0.421875\times 0.0625\\\\=0.263671875\\\\\approx 0.2637[/tex]
Thus, the probability that exactly 3 of the observed trucks are speeding is 0.2637.
(e)
Compute the probability that less than 3 of the observed trucks are speeding as follows:
[tex]P(X<3)=P(X=0)+P(X=1)+P(X=2)\\\\=\sum\limits^{2}_{x=0}{{5\choose x}(0.75)^{x}(1-0.75)^{5-x}}\\\\=0.0009765625+0.0146484375+0.087890625\\\\=0.103515625\\\\\approx 0.1035[/tex]
Thus, the probability that less than 3 of the observed trucks are speeding is 0.1035.
