Math- statistics - need help with problem 1-27(a-b) and 1-28(a-c) - please fully

Question

Math- statistics
- need help with problem 1-27(a-b) and 1-28(a-c)
- please fully explain the answer to following question
- thank you

Additional Problems scores on his first nine assignments are: 10, 10, 9, 9, 10, 8, 9, 10, and & a. What are th a. What is his mean (average) score so far? b. Mar e median, mode, and range of his scores? k did not do the tenth assignment for these assignments. What is his new mean? , so he got a zero on it. Zero is an 1-28. There are five students of different ages whose median age is 13 years. a. What are two possibilities for the ages of the three oldest students? b. If each student is a different age, how many students must be younger than 13? In complete sentences, describe what a median is to another student in the class. Pretend that the student was absent the day you learned about medians. c.

Explanation / Answer

1 - 27

(a)

Median:

Arranging numbers in ascending order: 8,8,9,9,9,10,10,10,10

n = total numbers = 9

Medin = (n+1)/2 th item = 5th item = 9

So, median = 9

Mode:

Number which occurs maximum number of times.

Here, 10 occurs maximum number of times.

So, mode = 10

Rage = maximum - minimum = 10 - 8 = 2

So,

Range = 2

(a) mean = total of all numbers/Total numbers = 83/9 = 9.2222

(b) new mean = 83/10 = 8.3

1 - 28

(a) Given: median = 13

Possiblity 1: 13,14,15

Possibility 2: 13,16,18

Note: Since 13 is the median, 3rd old student should be 13. Other 2 numbers can be anything greater than 13.

(b) Number of students younger than 13 is 2.

REASON: There are 5 students. Median is the 3nd student.

(c) When you arrange the numbers in ascending order, the middlemost item is the median/

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---