A machine fills containers with a particular product. The standard deviation of filling weights is known from past data to be 0.6 ounce. If only 2% of the containers hold less than 18 ounces, what is the mean filling weight for the machine? That is what must mu equal? Assume the filling weights have a normal distribution.

Diana Suarez

Diana Suarez

Answered question

2022-09-22

A machine fills containers with a particular product. The standard deviation of filling weights is known from past data to be 0.6 ounce. If only 2 % of the containers hold less than 18 ounces, what is the mean filling weight for the machine? That is what must μ equal? Assume the filling weights have a normal distribution.

Answer & Explanation

Brendon Melton

Brendon Melton

Beginner2022-09-23Added 5 answers

Step 1
let X denotes the fillings weights. the stanadrd deviation is given as 0.6 so the variance is 0.36
X N ( μ ,   0.36 )
given,
P [ X 18 ] = 0.02 P [ X μ 0.6 18 μ 0.8 ] = 0.02 Φ [ 18 μ 0.6 ] = 0.02 Φ [ 18 μ 0.6 ] = 1 0.88 Φ [ 18 μ 0.6 ] = 1 Φ [ 1.18 ] Φ [ 18 μ 0.6 ] = Φ [ 1.18 ] 18 μ 0.6 = 1.18 μ = 18 + 1.18 × 0.6 = 18.708   ounces

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