Respuesta :
Answer:
a) [tex]\bar X = 369.62[/tex]
b) [tex]Median=175 [/tex]
c) [tex] Mode =450 [/tex]
With a frequency of 4
d) [tex] MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5[/tex]
e)[tex] s = 621.76[/tex]
And we can find the limits without any outliers using two deviations from the mean and we got:
[tex] \bar X+2\sigma = 369.62 +2*621.76 = 1361[/tex]
And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
Step-by-step explanation:
We have the following data set given:
49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000
Part a
The mean can be calculated with this formula:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
Replacing we got:
[tex]\bar X = 369.62[/tex]
Part b
Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:
[tex]Median=175 [/tex]
Part c
The mode is the most repeated value in the sample and for this case is:
[tex] Mode =450 [/tex]
With a frequency of 4
Part d
The midrange for this case is defined as:
[tex] MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5[/tex]
Part e
For this case we can calculate the deviation given by:
[tex] s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
And replacing we got:
[tex] s = 621.76[/tex]
And we can find the limits without any outliers using two deviations from the mean and we got:
[tex] \bar X+2\sigma = 369.62 +2*621.76 = 1361[/tex]
And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case