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Sagot :
Answer:
The answer is "717.25 minutes".
Step-by-step explanation:
In this question first, we arrange the value in ascending order that are:
307,311,321,322,354,363,377,406,425,435,461,464,474,499,505,513,517,534,548
The median is always in an orderly position that is [tex]= \frac{(n+1)}{2}[/tex].
[tex]\to \frac{(n+1)}{2} = \frac{(20+1)}{2} = 10.5[/tex] position orders
[tex]10^{th} \ and \ 11^{th}[/tex] place an average of observations so, the
[tex]Average = \frac{(425 +435)}{2} = \frac{(860)}{2} =430[/tex]
Because as medium stands at 10.5 that median is as below 10 are greater than the level.
[tex]Q_1[/tex] often falls throughout the ordered BELOW average is [tex]= \frac{(n+1)}{2}[/tex] place.
[tex]\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}[/tex]ordered position
Average [tex]5^{th} \ and \ 6^{th}[/tex] location findings
[tex]Q_1 = \frac{(354+363)}{2} = \frac{(717)}{2}= 358.5[/tex]
In [tex]= \frac{(n+1)}{2}[/tex] the ordered place, Q3 always falls ABOVE the median.
[tex]\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}[/tex] ordered At median ABOVE
Consequently [tex]Q_3[/tex] falls between [tex]5^{th} \ and \ 6^{th}[/tex] ABOVE the median position
[tex]15^{th}\ and \ 16^{th}[/tex]place average of observations
[tex]\to Q_3 = \frac{(499+505)}{2}=\frac{(1004)}{2} = 502[/tex]
[tex]\to IQR = Q_3 - Q_1= 502-358.5= 143.5[/tex]
[tex]\to \text{Upper fence} = Q_3 + 1.5 \times IQR= 502 + 1.5 \times 143.5 = 717.25[/tex]
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