# Calculate Q1, D6, and P15 for the following data: Mid value 25 75 125 175 225 275 Frequency 10 70 80 100 150 90 - Mathematics and Statistics

Sum

Calculate Q1, D6, and P15 for the following data:

 Mid value 25 75 125 175 225 275 Frequency 10 70 80 100 150 90

#### Solution

Since the difference between any two consecutive mid values is 50, the width of each class interval is 50.
∴ the class intervals will be 0 – 50, 50 – 100, etc.
We construct the less than cumulative frequency table as given below:

 Class interval Frequency (f) Less than cumulative frequency (c.f.) 0 – 50 10 10 50 – 100 70 80 ← P15 100 – 150 80 160 ← Q1 150 – 200 100 260 200 – 250 150 410 ← D6 250 – 300 90 500 Total 500

Here, N = 500

Q1 class = class containing ("N"/4)^"th" observation

∴ "N"/4 = 500/4 = 125
Cumulative frequency which is just greater than (or equal) to 125 is 160.
∴ Q1 lies in the class 100 – 150.
∴ L = 100, h = 50, f = 80, c.f. = 80

∴ Q1 = "L"+"h"/"f"("N"/4-"c.f.")

= 100 + 50/80 (125 - 80)

= 100 + 5/8 (45)

= 100 + 28.125
= 128.125

D6 class = class containing ((6"N")/10)^"th" observation

∴ (6"N")/10=(6xx500)/10 = 300

Cumulative frequency which is just greater than (or equal) to 300 is 410.
∴ D6 lies in the class 200 – 250
∴ L = 200, h = 50, f = 150, c.f. = 260

∴ D6 = "L"+"h"/"f"((6"N")/4-"c.f.")

= 200 + 50/150 (300 - 260)

= 200 + 1/3 (40)

= 200 + 13.33
= 213.33

∴ P15 class = class containing ((15"N")/100)^"th" observation

∴ (15"N")/100 =(15 xx 500)/100 = 75
Cumulative frequency which is just greater than (or equal) to 75 is 80.
∴ P15 lies in the class 50 – 100
∴ L = 50, h = 50, f = 70, c.f. = 10

∴ P15 = "L"+"h"/"f"((15"N")/100-"c.f.")

= 50 + 50/70 (75 - 10)

= 50 + 5/7 (65)

= 50 + 325/7

= 50 + 46.4286
= 96.4286
∴ Q1 = 128.125, D6 = 213.33, P15 = 96.4286

Concept: Relations Among Quartiles, Deciles and Percentiles
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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board
Chapter 1 Partition Values
Miscellaneous Exercise 1 | Q 5 | Page 21