Question

A bottling machine can be regulated so that it discharges an average of μ ounces per bottle. It has been observed that the amount of fill dispensed by the machine has a normal distribution A sample of n= 16 filled bottles is randomly selected from the output of the machine on a given day and the ounces of fill measured for each. The sample variance is equal to one ounce. Find the probability: a) that each bottle filled will be within 0.3 ounce of the true mean? b) that the sample mean will be within 0.3 ounce of the true mean?

Answer 1

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 1
standard Deviation ( sd )= 1/ Sqrt ( 16 ) =0.25
sample size (n) = 16

a.
the probability that each bottle filled will be within 0.3 ounce of the true mean
P(X < 0.3) = (0.3-1)/1/ Sqrt ( 16 )
= -0.7/0.25= -2.8
= P ( Z <-2.8) From Standard NOrmal Table
= 0.00256

b.
the probability that the sample mean will be within 0.3 ounce of the true mean
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 1
standard Deviation ( sd )= 1
the probability that the sample mean will be within 0.3 ounce of the true mean
P(X < 0.3) = (0.3-1)/1
= -0.7/1= -0.7
= P ( Z <-0.7) From Standard Normal Table
= 0.242