Given the following data set: 0.23, 0.31, 0.33, 0.41, 0.56, 0.57, 0.75, 0.81 (a)

Question

Given the following data set: 0.23, 0.31, 0.33, 0.41, 0.56, 0.57, 0.75, 0.81 (a) Find the lower and upper quartiles. (Enter your answers to four decimal places.) Q1 = Q3 = (b) Calculate the IQR. (Enter your answer to four decimal places.) IQR = (c) Calculate the lower and upper fences. (Enter your answers to five decimal places.) lower fence upper fence Are there any outliers? Yes, 0.23 is an outlier. Yes, 0.81 is an outlier. Yes, 0.23 and 0.81 are outliers. Yes, 0.23 and 0.31 are outliers. No, there are no data points that lie outside these fences.

Explanation / Answer

0.23, 0.31, 0.33, 0.41, 0.56, 0.57, 0.75, 0.81

Here we have been given the sample size as n = 8,

a) The first quartile is computed as the 0.25(n+1)= 2.25th value of the observation

Now as the second value here is 0.31 and the third value is 0.33, therefore we get:

Q1 = 0.31 + 0.25*(0.33 - 0.31 ) = 0.315

Therefore Q1 = 0.315

The third quartile value is computed as the 0.75(n+1) = 6.75th value of the observation.

Now as the 6th value here is 0.57 and the 7th value is 0.75, therefore we get:

Q3 = 0.57 + 0.75*(0.75 - 0.57) = 0.705

Therefore we get: Q3 = 0.705

b) The interquartile range is computed as:

IQR = Q3 - Q1 =0.705 - 0.315 = 0.39

c) The lower and upper fence is computed as:

Lower fence = Q1 - 1.5*IQR = 0.315 - 1.5*0.39 = -0.27

Upper Fence = Q3 + 1.5*IQR = 0.705 + 1.5*0.39 = 1.29

d) Clearly as all the values lies within the lower and the upper fence, therefore No, there are no data points that lie outside these fences.

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