Find the median, first quartile, third quartile, interquartile range, and any outliers for each set of data. 111, 68, 93, 88, 74, 152, 119, 87, 88, 105, 84, 102, 151, 115, 112 A. Median: 102, Q1: 87, Q3: 113, IR: 28, outliers: none B. Median: 102, Q1: 87, Q3: 115, IR: 28, outliers: 68 C. Median: 102, Q1: 87, Q2: 115, IR: 28, outliers: none D. Median: 102, Q1: 87, Q3: 115, IR: 27, outliers: none

Q1 = 87 ;

Q2 = 102 ;

Q3 = 115 ;

IQR = 28

OUTLIER = None

Step-by-step explanation:

Given the data:

111, 68, 93, 88, 74, 152, 119, 87, 88, 105, 84, 102, 151, 115, 112

Ordered data : 68, 74, 84, 87, 88, 88, 93, 102, 105, 111, 112, 115, 119, 151, 152

Sample size, n = 15

The first quartile, Q1 = 1/4(n+1)th term

Q1 = 1/4(15+1)th term

Q1 = 1/4(16) = 4th term

Q1 = 87

The Median , Q2 = 1/2(n+1)th term

Q2 = 1/2(15+1)th term

Q2 = 1/2(16) = 8th term

Q2 = 102

The third quartile, Q3 = 3/4(n+1)th term

Q3 = 1/4(15+1)th term

Q3 = 3/4(16) = 12th term

Q3 = 115

The interquartile range, IQR = Q3 - Q1

IQR = (115 - 87) = 28

Q1 = 87 ;

Q2 = 102 ;

Q3 = 115 ;

IQR = 28

OUTLIER = None

b: |12 – x| ≤ 0.25

step-by-step explanation:      

let x be the possible lengths in inches, of the ribbons jaime cut.

we have been given that jaime is cutting strips of ribbon for a project. he tries to cut each piece to be 12 inches long, but sometimes he cuts it too long or too short by as much as 0.25 inch.  

we can represent this information in an inequality as:

by the definition of absolute value function we will get,

upon combining our inequalities we will get,

therefore, the inequality   describes all the possible lengths (x), in inches, of the ribbons jaime cut.

step-by-step explanation:

54, jfvhgnjhvnjhcbnncjigjm