Answer:
(115.2642, 222.7358).
Step-by-step explanation:
Given data:
type A: n_1=60, xbar_1=1827, s_1=168
type B: n_2=180, xbar_2=1658, s_2=225
n_1 = sample size 1, n_2= sample size 2
xbar_1, xbar_2 are mean life of sample 1 and 2 respectively. Similarly, s_1 and s_2 are standard deviation of 1,2.
a=0.05, |Z(0.025)|=1.96 (from the standard normal table)
So 95% CI is
(xbar_1 -xbar_2) ± Z×√[s1^2/n1 + s2^2/n2]
=(1827-1658) ± 1.96×sqrt(168^2/60 + 225^2/180)
= (115.2642, 222.7358).
The answer to the question is 2:3
<u>Given</u>:
Given that the data are represented by the box plot.
We need to determine the range and interquartile range.
<u>Range:</u>
The range of the data is the difference between the highest and the lowest value in the given set of data.
From the box plot, the highest value is 30 and the lowest value is 15.
Thus, the range of the data is given by
Range = Highest value - Lowest value
Range = 30 - 15 = 15
Thus, the range of the data is 15.
<u>Interquartile range:</u>
The interquartile range is the difference between the ends of the box in the box plot.
Thus, the interquartile range is given by
Interquartile range = 27 - 18 = 9
Thus, the interquartile range is 9.
Answer:
600
Step-by-step explanation:
36 times 18=648 but u need to add o the nearest hundred
Answer:
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