User32432432
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#1
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#1
Hey guys
S1 is just around the corner now and I've been doing past papers and I always see questions that say 'use linear interpolation formula to find the median' and its usually for 2 marks. I know the comparison method but I really dont like it and takes too much time, is there an alternate way ? maybe some sort of formula ?


Thanks
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Lunch_Box
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#2
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start of group + [(amount that median is into group)/frequency] * group width
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User32432432
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#3
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(Original post by Lunch_Box)
start of group + [(amount that median is into group)/frequency] * group width
Amount of median ?
I saw this b + <(0.5n - f)/ f > * C
but dont know what the letters represent and what 0.5 is
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JackClarke
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#4
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Median is Q2 so do n/2 and round up if it's not a whole number. (Where n = total freq) Then use interpolation.

Let's use some example data.

class | frequency
10 < a < 20 | 20
20 < a < 30 | 30
30 < a < 40 | 40

Total freq = n = 20+30+40 = 70

n/2 = 35

We are looking for the 35th value. This value clearly lies within 20 < a < 30 and we know we are looking for the (35-20 =) 15th value into this class.

Calculate the class width. In this case it is 30-20 = 10. (if the classes were 10-19, 20-29 then the class width is 29.5-19.5)

Now do class width/freq = 10/30 = 1/3

Multiply this by the number into the class.

1/3 * 15 = 15/3 = 5

Now do lower bound + 5 = 20+5 = 25.

This leaves us with: Lower bound of class + [class width/frequency] * number into the class
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User32432432
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#5
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(Original post by JackClarke)
Median is Q2 so do n/2 and round up if it's not a whole number. (Where n = total freq) Then use interpolation.

Let's use some example data.

class | frequency
10 < a < 20 | 20
20 < a < 30 | 30
30 < a < 40 | 40

Total freq = n = 20+30+40 = 70

n/2 = 35

We are looking for the 35th value. This value clearly lies within 20 < a < 30 and we know we are looking for the (35-20 =) 15th value into this class.

Calculate the class width. In this case it is 30-20 = 10. (if the classes were 10-19, 20-29 then the class width is 29.5-19.5)

Now do class width/freq = 10/30 = 1/3

Multiply this by the number into the class.

1/3 * 15 = 15/3 = 5

Now do lower bound + 5 = 20+5 = 25.

This leaves us with: Lower bound of class + [class width/frequency] * number into the class
Thank you
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User32432432
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#6
Report Thread starter 6 years ago
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(Original post by JackClarke)
Median is Q2 so do n/2 and round up if it's not a whole number. (Where n = total freq) Then use interpolation.

Let's use some example data.

class | frequency
10 < a < 20 | 20
20 < a < 30 | 30
30 < a < 40 | 40

Total freq = n = 20+30+40 = 70

n/2 = 35

We are looking for the 35th value. This value clearly lies within 20 < a < 30 and we know we are looking for the (35-20 =) 15th value into this class.

Calculate the class width. In this case it is 30-20 = 10. (if the classes were 10-19, 20-29 then the class width is 29.5-19.5)

Now do class width/freq = 10/30 = 1/3

Multiply this by the number into the class.

1/3 * 15 = 15/3 = 5

Now do lower bound + 5 = 20+5 = 25.

This leaves us with: Lower bound of class + [class width/frequency] * number into the class
Are there any videos on youtube which may help me understand it better ?
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revisor1233333
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#7
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#7
Estimated Median = L + (n/2) − B × w
G
where:

L is the lower class boundary of the group containing the median
n is the total number of values
B is the cumulative frequency of the groups before the median group
G is the frequency of the median group
w is the group width
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17011302
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#8
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0.5 is just showing its a median, if you were finding lower quartile it would be 0.25, 63rd percentile 0.63 etc.For a median:Lower class boundary ((n/2 - frequency of group your in) ÷ cumulative frequency of the previous groups) * class width
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