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Statistical Analysis of Percentage of Underfilled Bottles in a Production Process, Exercises of Statistics

Statistical calculations of the percentage of underfilled bottles based on given mean and standard deviation values. It includes calculations for different standard deviations and percentages of underfilled bottles, as well as the fill amount exceeded by certain percentages of bottles.

Typology: Exercises

2019/2020

Uploaded on 02/23/2020

het-verugamwala
het-verugamwala 🇨🇦

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a) µ = 302 = 4, % under filled bottles = 1 - P(X > 300) x 100 = 30.85%
µ = 304 = 4, % under filled bottles = 1 P(X > 300) x 100 = 15.87%
b) µ = 302 = 6, % under filled bottles = 1 - P(X > 300) x 100 = (1 0.6306) x 100 = 36.94%
c) µ = 302 = 4, % under filled bottles within 1 standard deviation = P (298<= X <= 306) x 100
=68.27%
µ = 302 = 4, % under filled bottles within 2 standard deviation = P (294<= X <= 310) x 100
= 95.45%
d) = 302 ml and = 4 ml, the fill amount exceeded by 95% of the bottles = 295.42 mL
= 302 ml and = 4 ml, the fill amount exceeded by 99% of the bottles = 292.70 mL
e) The mean amount () in order that only 5% of the bottles receive less than 296 ml = 302.58 mL
0.3085
0.1587
0.3694
0.6827
0.9545
295.42 mL
292.70 mL
302.58 mL

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a) μ = 302  = 4, % under filled bottles = 1 - P(X > 300) x 100 = 30.85% μ = 304  = 4, % under filled bottles = 1 – P(X > 300) x 100 = 15.87% b) μ = 302  = 6, % under filled bottles = 1 - P(X > 300) x 100 = (1 – 0.6306) x 100 = 36.94% c) μ = 302  = 4, % under filled bottles within 1 standard deviation = P (298<= X <= 306) x 100 =68.27% μ = 302  = 4, % under filled bottles within 2 standard deviation = P (294<= X <= 310) x 100 = 95.45% d)  = 302 ml and  = 4 ml, the fill amount exceeded by 95% of the bottles = 295.42 mL  = 302 ml and  = 4 ml, the fill amount exceeded by 99% of the bottles = 292. 70 mL e) The mean amount () in order that only 5% of the bottles receive less than 296 ml = 302. 58 mL

295.42 mL 292.70 mL 302.58 mL