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

Answer:

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

Step-by-step 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

              $\sigma_{\= x } = \frac{\sigma}{\sqrt{n} }$

substituting values

            $\sigma_{\= x } = \frac{0.1}{\sqrt{6} }$

            $\sigma_{\= x } = 0.0408$

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

         $P(X < 12) = P ( \frac{X - \mu }{ \sigma_{\= x }} < \frac{\= x - \mu }{ \sigma_{\= x }} )$

Generally  $\frac{X - \mu }{ \sigma_{ \= x }} = Z (The \ standardized \ value \ of \ X )$

So

         $P(X < 12) = P ( Z < \frac{\= x - \mu }{ \sigma_{\= x }} )$

substituting values

       $P(X < 12) = P ( Z < \frac{12.2 -12 }{0.0408} )$

      $P(X < 12) = P ( Z < 2.45 )$

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

           $P (Z < 2.45)0.99286$

=>   $P(X < 12) = 0.99286$