rmal Distribution of the Sample Mean Soft drink cans are filled by an automated filling machine.

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answerhappygod
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rmal Distribution of the Sample Mean Soft drink cans are filled by an automated filling machine.

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Normal Distribution of the Sample Mean Soft drink cans are filled by an automated filling machine. The mean fill volume is 12.1 fluid Ounces, and the standard deviation is 0.05 fluid Ounces. Assume that the fill volumes of the cans are independent, random normal variables. a) What is the probability that the average volume of 10 can, selected from this process is less than 12 fluid Ounces? b) What is the probability that the average volume of 10 can, selected from this process is between 11 and 13 fluid Ounces?


We have :

μ = 12.1 and σ = 0.05

x = volume of can : follow normal distribution

x̄ = sample mean or average follow normal distribution with :

μ x̄ = μ and σ x̄ = σ / sqrt ( n) = 0.0158

here n = 10 can

a ) Probability that the average volume of 10 can selected from this process is less than 12 fluid ounces

= P [ x̄ < 12 ]

we know that , z = ( x̄ - μ ) / σ x̄ follow standard normal distribution

z value when x̄ = 12

z = ( 12 - 12.1 ) / 0.0158

= - 6.3246

P [ z < - 6.3246 ] = 1 - P [ z < 6.3246 ]

= 1 - 1

= 0

( 0 is the Probability that the average volume of 10 can selected from

this process is less than 12 fluid ounces )

## b )

Probability that the average volume of 10 can selected from this process is between 11 fluid and

13 fluid vlolumes

= P [ 11 < x̄ < 13 ]

we know that , z = ( x̄ - μ ) / σ x̄ follow standard normal distribution

z value when x̄ = 11

z = ( 11 - 12.1 ) / 0.0158

= - 69.5701

z value when x̄ = 13

z = ( 13 - 12.1 ) / 0.0158

= 56.9201

P [ 11 < x̄ < 13 ] = P [ - 69.5701 < z < 56.9201 ]

= P [ z < 56.9201 ] - P [ z < - 69.5701 ]

= P [ z < 56.9201 ] - ( 1 - P [ z < - 69.5701 ] )

= 1 - ( 1 - 1 )

= 1 - 0

= 1

( 1 is Probability that the average volume of 10 can selected from

this process is between 11 fluid and 13 fluid vlolumes )
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