8, 5, 6, 16, 22, 13, 24, 2 Median -

Q1 -

Q3 -

IQR -

63, 45, 21, 14, 7, 3, 1

Median -

Q1 -

Q3 -

IQR -

Median #1: 10.5

Q1 #1: 5.5

Q3 #1: 19

IQR #1: 13.5

Median #2: 14

Q1 #2: 3

Q3 #2: 45

IQR #2: 42

Step-by-step explanation:

Question 1:

First order the data so that they are in order from least to greatest:

2, 5, 6, 8, 13, 16, 22, 24

Now, find the median by crossing out one value on the left and one value on the right until you get to the center value.

2, 5, 6, 8, 13, 16, 22, 24

/2/, /5/, /6/, /8/, /13/, /16/, /22/, /24/

In this case, there were four values crossed off of each side, so you'll need to find the average of the final two crossed out values: 8 and 13.

(8+13)/2 = 10.5

To find the first quartile, find the median of all values below the median.

2, 5, 6, 8

/2/, /5/, /6/, /8/

The median of these values is the average of the center two, so

(5+6)/2 = 5.5

The third quartile is the median of all values above the median.

13, 16, 22, 24

/13/, /16/, /22/, /24/

The median of those values is the average of the center two as well, so

(16 + 22)/2 = 19

The IQR can be found by subtracting Q1 from Q3. (IQR = Q3 - Q1)

Q3 - Q1

19 - 5.5

IQR = 13.5

Question 2:

Arrange the values from least to greatest and cross them out until you find the middle falue, like in problem one.

1, 3, 7, 14, 21, 45, 63

/1/, /3/, /7/, 14, /21/, /45/, /63/

The center value is 14, so the median is 14.

The first quartile can be found by finding the median of all values below the median.

1, 3, 7

/1/, 3, /7/

The first quartile is 3.

The third quartile can be found by finding the median of all values above the median.

21, 45, 63

/21/, 45, /63/

The third quartile is 45.

The interquartile range can be found using the equation from above.

Q3 - Q1

45 - 3

The interquartile range is 42.

the answer is a) y= 2 csc [0.5 (x+pi/2)]-1

step-by-step explanation: