Question

Sodas in a can are supposed to contain an average 12 oz. This particular brand has a standard deviation of 0.1 oz, with an average of 12.1 oz. If the can’s contents follow a Normal distribution, what is the probability that the mean contents of a six-pack are less than 12 oz?

Answer 1

The probability is
`P(X < 12) = 0.99286`

Explanation:

From the question we are told that

The population mean is
`\mu = 12 \ oz`

The standard deviation is
`\sigma = 0.1 \ oz`

The sample mean is
`\= x = 12.1 \ oz`

The sample size is n = 6 packs

The standard error of the mean is mathematically represented as

`\sigma_(\= x ) = (\sigma)/(√(n) )`

substituting values

`\sigma_(\= x ) = (0.1)/(√(6) )`

`\sigma_(\= x ) = 0.0408`

Given that the can’s contents follow a Normal distribution then then the probability that the mean contents of a six-pack are less than 12 oz is mathematically represented as

`P(X < 12) = P ( (X - \mu )/( \sigma_(\= x )) < (\= x - \mu )/( \sigma_(\= x )) )`

Generally
`(X - \mu )/( \sigma_( \= x )) = Z (The \ standardized \ value \ of \ X )`

So

`P(X < 12) = P ( Z < (\= x - \mu )/( \sigma_(\= x )) )`

substituting values

`P(X < 12) = P ( Z < (12.2 -12 )/(0.0408) )`

`P(X < 12) = P ( Z < 2.45 )`

From the normal distribution table the value of
`P ( Z < 2.45 )` is

`P (Z < 2.45)0.99286`

=>
`P(X < 12) = 0.99286`