Answer:
This is the answer and I hope it helps. have a great time.
Answer:
0.9179 = 91.79% probability that a randomly selected call time will be less than 25 seconds
Step-by-step explanation:
Exponential distribution:
The exponential probability distribution, with mean m, is described by the following equation:
In which 
is the decay parameter.
The probability that x is lower or equal to a is given by:
Which has the following solution:
The probability of finding a value higher than x is:
In this question:
Find the probability that a randomly selected call time will be less than 25 seconds?
0.9179 = 91.79% probability that a randomly selected call time will be less than 25 seconds
Answer:
7/3 or 13/15
Step-by-step explanation:
So you told me that it will be
8/(20/9) -2 11/15 I put parenthesis to make it easier to understand for me
so you must know that a/(b/c)=ac/b si
72/20 -2 11/15 so we work with -2 11/15 which is equal to -30/15 + 11/15 and that will be -19/15 so we have
72/20 - 19/15 we simplify
18/5 - 19/15 so we multiply and divide by 3 in 18/5
54/15 - 19/15 and we add up
35/15 which is equal to
7/3
If this is wrong <u>so we work with -2 11/15 which is equal to -30/15 + 11/15 </u> well it should be so we work with -2 11/15 which is equal to -30/15 - 11/15 and that equals -41/15 and we get
54/15-41/15
13/15
Answer:
0.184
Step-by-step explanation:
As the statistician consultant, I would have to calculate the chance of having 2 ambulances on the freeway at this hours and relay the message to the dispatcher.
Probability of having the need for 2 ambulances.
We will have a poisson distribution:
Lambda = 1
P(x=2) = e^-2*1/2!
= 2.71828^-1/2
= 0.368/2
= 0.184
I would tell the dispatch rider that the possibility that 2 ambulances would be required is 0.184.
Thank you
Answer:
The correct value of z crit for this test is +1.645
Step-by-step explanation:
