Question

grade 6 unit 1
georgias k - 12 mathematics standards
using the data from set a and set b, compare the following:
34, 40, 44, 52, 55, 57, 58, 60, 64, 71
30, 32, 48, 50, 53, 57, 57, 58, 62, 68
median set a set b
minimum 55 53
maximum
first quartile
third quartile
range
interquartile range

Explanation:

Step1: Find minimum of Set A

The minimum is the smallest value. For Set A: 34, 40, 44, 52, 55, 57, 58, 60, 64, 71, the minimum is 34.

Step2: Find minimum of Set B

For Set B: 30, 32, 48, 50, 53, 57, 57, 58, 62, 68, the minimum is 30.

Step3: Find maximum of Set A

The maximum is the largest value. For Set A, the maximum is 71.

Step4: Find maximum of Set B

For Set B, the maximum is 68.

Step5: Find first - quartile of Set A

There are n = 10 data points. The position of the first - quartile ($Q_1$) is $L_{Q_1}=\frac{n + 1}{4}=\frac{10+1}{4}=2.75$. So, $Q_1=40+(44 - 40)\times0.75 = 43$.

Step6: Find first - quartile of Set B

For n = 10 data points, $L_{Q_1}=\frac{n + 1}{4}=2.75$. So, $Q_1=32+(48 - 32)\times0.75=44$.

Step7: Find third - quartile of Set A

The position of the third - quartile ($Q_3$) is $L_{Q_3}=\frac{3(n + 1)}{4}=\frac{3\times(10 + 1)}{4}=8.25$. So, $Q_3=60+(64 - 60)\times0.25 = 61$.

Step8: Find third - quartile of Set B

For n = 10 data points, $L_{Q_3}=\frac{3(n + 1)}{4}=8.25$. So, $Q_3=58+(62 - 58)\times0.25 = 59$.

Step9: Find range of Set A

Range = Maximum - Minimum. For Set A, Range = 71 - 34 = 37.

Step10: Find range of Set B

For Set B, Range = 68 - 30 = 38.

Step11: Find inter - quartile range of Set A

Inter - quartile range (IQR)=$Q_3 - Q_1$. For Set A, IQR = 61 - 43 = 18.

Step12: Find inter - quartile range of Set B

For Set B, IQR = 59 - 44 = 15.

Answer:

	Set A	Set B
Minimum	34	30
Maximum	71	68
First quartile	43	44
Third quartile	61	59
Range	37	38
Interquartile Range	18	15