Answer to Question #324147 in Statistics and Probability for pnkj

Question

A machine fills coffee powder in pouches, with an average of 200 gm and a standard deviation of 4 gm. Assuming that the coffee weight is normally distributed. Find the probability that a coffee pouch selected at random will contain the following quantity of a coffee: I) At least 200 gm. II) Between 200 to 206 gm

Answer 1

mean($\bar x$ )=200

standard deviation (s)=4

z=$\frac{x-\bar x}{s}$

1.

x=200

P(x$\geq200)=1-P(x<200)$

z=$\frac{200-200}{4}$ =0

P(z<0)=0.5

P(x$\geq$ 200)=1-0.5

P(x$\geq$ 200)=0.5

The probability that a coffee pouch selected at random will contain at least 200 gm is 0.5.

2. Between 200 and 206 .

P(200<x<206)= P(z₁<x<z₂)

z₂=$\frac{206-200}{4}$

z₂=1.5

P(z₂$\le$ 1.5)=0.9332

using z₁ previously computed

P(z₁$\le$0)=0.5

P(z₁<x<z₂)=0.9332-0.5

P(z₁<x<z₂)=0.4332

The probability that a coffee pouch selected at random will contain between 200 gm and 206 gm is 0.4332.