Find the length of arc JK. (Round to the nearest hundredth).
2 answers:
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The box-and-whisker plot is attached.
We first order the data from least to greatest:
67,68,68,69,70,72,73,74,75,75,76,78,79,79,80,83,84,85,85,88
Now we find the median. Since there are 20 data values, the median is between 75 and 76:
(75+76)/2 = 151/2 = 75.5
Now we find the Lower Quartile (LQ). We do this by finding the median of the lower half of data (from the median down). There are 10 values here; the median is between 70 and 72:
(70+72)/2 = 142/2 = 71
Find the Upper Quartile (UQ). Do this by finding the median of the upper half of data (from the median up). There are 10 values here; the median is between 80 and 83:
(80+83)/2 = 163/2 = 81.5
The middle bar of the box is over the median. The left hand side of the box is over LQ, and the right hand side of the box is over UQ. Now we draw a whisker from the box to the highest value, 88, and from the box to the lowest value, 67.
We first order the data from least to greatest:
67,68,68,69,70,72,73,74,75,75,76,78,79,79,80,83,84,85,85,88
Now we find the median. Since there are 20 data values, the median is between 75 and 76:
(75+76)/2 = 151/2 = 75.5
Now we find the Lower Quartile (LQ). We do this by finding the median of the lower half of data (from the median down). There are 10 values here; the median is between 70 and 72:
(70+72)/2 = 142/2 = 71
Find the Upper Quartile (UQ). Do this by finding the median of the upper half of data (from the median up). There are 10 values here; the median is between 80 and 83:
(80+83)/2 = 163/2 = 81.5
The middle bar of the box is over the median. The left hand side of the box is over LQ, and the right hand side of the box is over UQ. Now we draw a whisker from the box to the highest value, 88, and from the box to the lowest value, 67.
Answer:
CI = (98.11 , 98.49)
The value of 98.6°F suggests that this is significantly higher
Step-by-step explanation:
Data provided in the question:
sample size, n = 103
Mean temperature, μ = 98.3
°
Standard deviation, σ = 0.73
Degrees of freedom, df = n - 1 = 102
Now,
For Confidence level of 99%, and df = 102, the t-value = 2.62 [from the standard t table]
Therefore,
CI =
Thus,
Lower limit of CI =
or
Lower limit of CI =
or
Lower limit of CI = 98.11
and,
Upper limit of CI =
or
Upper limit of CI =
or
Upper limit of CI = 98.49
Hence,
CI = (98.11 , 98.49)
The value of 98.6°F suggests that this is significantly higher and the mean temperature could very possibly be 98.6°F