Respuesta :
Answer: 0.2033
Step-by-step explanation:
Given : The departure time varies randomly, with expected departure times 12 o'clock noon and a standard deviation of 6 minutes.
Let [tex]\mu=0[/tex]
[tex]\sigma=6[/tex]
Let x be the random variable that represents the departure time .
We assume the departure time is normally distributed.
The formula to find the z-score :-
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x= 5 , we have
[tex]z=\dfrac{5-0}{6}\approx0.83[/tex]
Now, the probability that the bus has not yet departed :-
[tex]P(x>5)=P(z>0.83)=1-P(z\eq0.83)[/tex]
[tex]=1- 0.7967306=0.2032694\approx0.2033[/tex]
Hence, the probability that the bus has not yet departed = 0.2033
The value of the z-score is 0.83333. Then the probability that the bus has not yet departed is 0.2033 or 20.33 %.
What is normal a distribution?
It is also called the Gaussian Distribution. It is the most important continuous probability distribution. The curve looks like a bell, so it is also called a bell curve.
The z-score is a numerical measurement used in statistics of the value's relationship to the mean of a group of values, measured in terms of standards from the mean.
My bus is scheduled to depart at noon.
However, in reality, the departure time varies randomly, with expected departure times 12 o'clock noon and a standard deviation of 6 minutes.
Assume the departure time is normally distributed.
If I get to the bus stop 5 minutes past noon,
Let
μ = 0
σ = 6
X = 5
Then the z-score will be
[tex]\rm z-score = \dfrac{X - \mu }{\sigma}\\\\z-score = \dfrac{5-0}{6} \\\\z-score = 0.8333[/tex]
Then the probability of the bus has not yet departed will be
[tex]\rm P(X > 5) = P(z > 0.83) = 1-P(z 0.93)\\\\P(X > 5) = P(z > 0.83) = 1- 0.7967\\\\P(X > 5) = P(z > 0.83) = 0.2033[/tex]
More about the normal distribution link is given below.
https://brainly.com/question/12421652
