Professor Melendez has 10 students in her college
algebra class. Their ages are shown below.
When the outliers are removed, what is the mean
age of the remaining students? Round your
answer to the nearest tenth
19.27 19.18.18.18.7.19.2018
The mean age of the students in the class is 22.3.
The median age of the students is 19.
What is the median age of the remaining
students?
There are two ourers in the data from Professor
Melender's class: 27 and 47
DONE

Mean = 18.625

Median = 18.5

Step-by-step explanation:

Given the following :

Age = 19, 27, 19, 18, 18, 18, 47, 19, 20, 18

The mean Age :

Mean = Σ(x) ÷ N

N = number of data

(19 + 27 + 19 + 18 + 18 + 18 + 47 + 19 +20 + 18) / 10

Σ(x) / N = 223/10 = 22.3

Rearranging to get the median:

18,18,18,18,19,19,19,20,27,47

Middle value = median ( 19 + 19)/2 = 19

The outliers in the data are:

18,18,18,18,19,19,19,20,27,47

OUTLIERS ARE VALUES

< Q1 - (1.5 × IQR)

> Q3 + (1.5 × IQR)

IQR = Q3 - Q1 (Interquartile range)

Q3 = upper quartile

Q1 = lower quartile

From the data :

Q3 = 20, Q1 = 18

IQR = Q3 - Q1 = 20 - 18 = 2

< 18 - (1.5 × 2) ; <15

> 20 + (1.5 × 2) ; >23

VAlues greater than 23 and less than 15 are outliers in the data

27 and 47

After removing outliers

N = 10 - 2 = 8

Σ(x) = 223 - (47+27) = 149

Mean = 149/8 = 18.625

X = 18,18,18,18,19,19,19,20

Median = (18 + 19)/2 = 18.5

depends on how many days you are talking about

step-by-step explanation:

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.