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Question
The measurements of the diameters (in cms) of the plates prepared in a factory are given below. Find the standard deviation
| Diameter (cm) | 21 − 24 | 25 − 28 | 29 − 32 | 33 − 36 | 37 − 40 | 41 − 44 |
| Number of plates | 15 | 18 | 20 | 16 | 8 | 7 |
Solution
Assumed mean = 34.5
| Diameter (cm) | Number of plates fi |
mid value xi |
di = xi − A = xi − 34.5 |
fidi | fidi2 |
| 21 − 24 | 15 | 22.5 | − 12 | − 180 | 2160 |
| 25 − 28 | 18 | 26.5 | − 8 | − 144 | 1152 |
| 29 − 32 | 20 | 30.5 | − 4 | − 80 | 320 |
| 33 − 36 | 16 | 34.5 | 0 | 0 | 0 |
| 37 − 40 | 8 | 38.5 | 4 | 32 | 128 |
| 41 − 44 | 7 | 42.5 | 8 | 56 | 448 |
| `sumf_"i"` = 84 | `sumf_"i"d_"i"` = − 316 | `sumf_"i""d"_"i"^2` = 4208 |
Here `sumf_"i"` = 84, `sumf_"i"d_"i"` = − 316, `sumf_"i""d"_"i"^2` = 4208
N = `sumf_i = 84`
Standard deviation (σ) = `sqrt((sumf_"i""d"_"i"^2)/(sumf_"i") - ((sumf_"i""d"_"i")/(sumf_"i"))^2`
= `sqrt(4208/84 - (316/84)`
= `sqrt(50.095 - (3.76)^2`
= `sqrt(50.1 - 14.14)`
= `sqrt(35.96)`
= 5.996
= 6
∴ Standard deviation (σ) = 6
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