A lake near the Arctic Circle is covered by a sheet of ice during the cold winter months. When spring arrives, the ice starts to
2 answers:
Answer: 1.5 meters
Step-by-step explanation:
Given: A lake near the Arctic Circle is covered by a sheet of ice during the cold winter months. S
S(t) models the ice sheet's thickness (in meters) as a function of time t (in weeks)
if we put x=0, then
So 4 meters is the initial thickness
To find the thickness after 6 weeks, put x=6 we get
Decrease in thickness every 6 weeks=
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Answer:
a)
b)
c)
And we can find the usual limits with:
And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.
d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.
And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.
Step-by-step explanation:
We have the following data:
11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2
Part a
We can calculate the sample mean with the following formula:
And if we replace we got:
For the median we need to sort the values on increasing order and we have:
8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8
Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:
The mode would be the most repeated value on this case:
Part b
The range is defined as and if we replace we got:
We can calculate the standard deviation with the following formula:
And if we replace we got:
Part c
For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.
We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and
And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got
Then we can find the IQR like this:
And we can find the usual limits with:
And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.
Part d
The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.
And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.