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A company produces tables which are packed in batches of 100. An analysis of the defective tubes in different batches has received the following information: No. of defective tubes Less than 5 5 – 9 - Mathematics and Statistics

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प्रश्न

A company produces tables which are packed in batches of 100. An analysis of the defective tubes in different batches has received the following information:

No. of defective tubes Less than 5 5 – 9 10 – 14 15 – 9 20 – 24 25 – 29 30 and above
No. of tubes 45 51 84 39 20 8 4

Estimate the number of defective tubes in the central batch.

योग

उत्तर

To find the number of defective tubes in the central batch, we have to find Q2.
Since, the given data is not continuous, we have to convert it in the continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval.
∴ the class intervals will be
Less than 4.5, 4.5 – 9.5, etc.
We construct the less than cumulative frequency table as given below:

No. of defective tubes No. of tubes
(f)
Less than cumulative frequency
(c.f.)
Less than 4.5 45 45
4.5 – 9.5 51 96
9.5 – 14.5 84 180 ← Q2
14.5 – 19.5 39 219
19.5 – 24.5 20 239
24.5 – 29.5 8 247
29.5 and above 4 251
Total 251  

Here, N = 251
Q2 class = class containing `((2"N")/4)^"th"` observation

∴ `(2"N")/4=(2xx251)/4` = 125.5

Cumulative frequency which is just greater than (or equal to) 125.5 is 180.
∴ Q2 lies in the class 9.5 – 14.5
∴ L = 9.5, h = 5, f = 84, c.f. = 96

∴ Q2 = `"L" "h"/"f"((2"N")/4-"c.f.")`

= `9.5 + 5/84 (125.5 - 96)`

= `9.5 + 5/84 xx 29.5`

= `9.5 + (147.5)/84`

= 9.5 + 1.76
= 11.26

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Partition Values
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Partition Values - Miscellaneous Exercise 1 [पृष्ठ २२]

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बालभारती Mathematics and Statistics 2 (Commerce) [English] 11 Standard Maharashtra State Board
अध्याय 1 Partition Values
Miscellaneous Exercise 1 | Q 14 | पृष्ठ २२
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