0.35 cups/hour
To be able to determine the rate at which the water is leaking from the pipe with the information given, you have to divide the number of cups by the number of hours in which they were collected:
12 cups/34 hours= 0.35 cups/hour
According to this, the answer is that the rate at which the water is leaking from the pipe is 0.35 cups/hour.
Hope this helps!!
Answer:
Step-by-step explanation:
a )
sample mean = sum total of given data / no of data
= 415.35 / 20 = 20.76
To calculate the median we arrange the data in ascending order and take the average of 10 th and 11 th term .
= 20.50 + 20.72 / 2
= 20.61
b ) To calculate the 10% trimmed mean , we neglect the largest 10% and smallest 10 % data and then calculate the mean . Here we neglect the first two smallest and last two greatest
(18.92 + 19.25 ..... + 22.43 + 22.85) / 16
= 20.74
c )
We can easily plot the data on number line from 17 to 24
d )
Maximum value of data set = 23.71 and minimum value is 18.04
mean is 20.76 , median is 20.61 and trimmed mean is 20.74
They are between maximum and minimum values of given data . Hence there is no outliers .
It is -11.34
-$68.04 divide by 6 is -$11.34, so instead of 11.34, it's -11.34
(a) Data with the eight day's measurement.
Raw data: [60,58,64,64,68,50,57,82],
Sorted data: [50,57,58,60,64,64,68,82]
Sample size = 8 (even)
mean = 62.875
median = (60+64)/2 = 62
1st quartile = (57+58)/2 = 57.5
3rd quartile = (64+68)/2 = 66
IQR = 66 - 57.5 = 8.5
(b) Data without the eight day's measurement.
Raw data: [60,58,64,64,68,50,57]
Sorted data: [50,57,58,60,64,64,68]
Sample size = 7 (odd)
mean = 60.143
median = 60
1st quartile = 57
3rd quartile = 64
IQR = 64 -57 = 7
Answers:
1. The average is the same with or without the 8th day's data. FALSE
2. The median is the same with or without the 8th day's data. FALSE
3. The IQR decreases when the 8th day is included. FALSE
4. The IQR increases when the 8th day is included. TRUE
5. The median is higher when the 8th day is included. TRUE
