Respuesta :
Answer:
A)Mean :10.65
Median =10.7
Mode : 10.7
B)Range = 3.5
Standard deviation :0.89916
C)The response time of 8.3 minutes should be considered an outlier in comparison to the other response times
Step-by-step explanation:
A)
Data : 11.8 ,10.3, 10.7, 10.6, 11.5, 8.3, 10.5, 10.9, 10.7, 11.2
[tex]Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\Mean = \frac{11.8 +10.3+ 10.7+ 10.6+ 11.5+ 8.3+ 10.5+ 10.9+ 10.7+ 11.2}{10}[/tex]
Mean = 10.65
Median: The mid value of the data
Data in ascending order
8.3
10.3
10.5
10.6
10.7
10.7
10.9
11.2
11.5
11.8
n=10(even)
[tex]Median = \frac{(\frac{n}{2})\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}\\Median = \frac{(\frac{10}{2})\text{th term}+(\frac{10}{2}+1)\text{th term}}{2}\\Median = \frac{10.7+10.7}{2}\\Median = 10.7[/tex]
Mode : the most occurring frequency
10.7 is occurring twice while others are occurring once
So, Mode is 10.7
B) Range = Maximum - Minimum=11.8-8.3=3.5
Standard deviation : [tex]\sqrt{\frac{\sum(x-\bar{x})^2}{n}}[/tex]
Standard deviation :[tex]\sqrt{\frac{(11.8-10.65)^2+(10.3-10.65)^2+......+(10.9-10.65)^2+(10.7-10.65)^2+(11.2-10.65)^2}{10}}[/tex]
Standard deviation :0.89916
C)
8.3
,10.3
,10.5
,10.6
,10.7
,10.7
,10.9
,11.2
,11.5
,11.8
For Q1 ( Median of lower quartile )
8.3
,10.3
,10.5
,10.6
,10.7
[tex]Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=10.5[/tex]
For Q3( Median of Upper quartile )
10.7
,10.9
,11.2
,11.5
,11.8
[tex]Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=11.2[/tex]
IQR = Q3-Q1=11.2-10.5=0.7
Range :(Q1-1.5IQR, Q3-1.5IQR)
Range :[tex](10.5-1.5 \times 0.7, 11.2-1.5 \times 0.7)[/tex]
Range :(9.45, 10.15)
8.3 does not lie in this interval
So, the response time of 8.3 minutes should be considered an outlier in comparison to the other response times
