The average is calculated as 60 / 4 = 15
13 - 15 = -2
17 - 15 = 2
9 - 15 = -6
21 - 15 = 6
The total of squares amounts to 4 + 4 + 36 + 36 = 80
Since this is a sample, we divide it by n - 1 (which equals 3)
The standard deviation is calculated as sqrt(80/3) = 5.16 as the final answer.
Answer:
The likelihood that the number of drivers is at most 18 is 0.381.
Step-by-step explanation:
We are provided with the following details in the question:
The quantity of drivers traveling between a specific origin and destination in a set time frame follows a Poisson distribution characterized by the parameter μ = 20.
- The Poisson distribution defines the probability of a certain number of events taking place over a specific period, based on the mean frequency of those events.
- The variance for the Poisson distribution matches its mean value of Poisson distribution.
a) P(number of drivers will be at most 18)
Equation:


So, 0.381 represents the probability that the number of drivers will be at most 18.
The initial equation is

. The subsequent equation is

. By setting both equations equal, we can derive the following:


. Next, simplify:

. Then factor:

. Put x=0 or x=1 into the second equation to yield:

. Alternatively,

. Thus, the solutions are:

.
Answer:
Attached is the histogram illustrating the marathon runners’ times.
Step-by-step explanation:
The provided data is as follows;
2.21
2.25
2.76
3.1
3.3
3.5
3.6
3.77
3.8
4.23
4.25
4.25
4.6
4.9
From this data, we can determine;
The count of runners finishing between 0 and 1 hour = 0
The count of runners finishing between 1 and 2 hours = 0
The count of runners finishing between 2 and 3 hours = 3
The count of runners finishing between 3 and 4 hours = 6
The count of runners finishing between 4 and 5 hours = 5
Based on these frequencies across the various time ranges, the histogram for the provided data has been constructed and is attached.
To solve this problem, simply sum all of the grocery expenses and then divide that total by 4...