Respuesta :
Answer:
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
Answer:
Here's all the answer:
Professor Melendez has 10 students in her college algebra class. Their ages are shown below.
19, 27, 19, 18, 18, 18, 47, 19, 20, 18
The mean age of the students in the class is 22.3. The median age of the students is 19.
There are two outliers in the data from Professor Melendez’s class: 27 and 47.
When the outliers are removed, what is the mean age of the remaining students? Round your answer to the nearest tenth
⇒ 18.6
What is the median age of the remaining students?
⇒18.5
When outliers are removed from data, the greatest change generally occurs in the ✔ mean .
Step-by-step explanation:
got it from E D G E N U I T Y
hope this helps