Question

Bottles filled by a certain machine are supposed to contain 12 oz of liquid. In fact the fill volume is random with mean 12.01 oz and standard deviation 0.35 oz.a) What is the probability that the mean volume of a random sample of 144 bottles is less than 12 oz?b) If the population mean fill volume is increased to 12.03 oz, what is the probability that the mean volume of a sample of size 144 will be less than 12 oz?

Answer 1

Answer:

a

 $P(X < 12) = 0.3660$

b

  $P(X < 12) = 0.1521$

Step-by-step explanation:

Considering question a

From the question we are told that

   The mean is  $\mu = 12.01 \ oz$

    The standard deviation is  $\sigma = 0.35 \ oz$

 The standard error of mean is mathematically represented as

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

=>   $\sigma_{x} = \frac{0.35}{\sqrt{144} }$              

=>   $\sigma_{x} = 0.02917$

Generally the  probability that the mean volume of a random sample of 144 bottles is less than 12 oz is mathematically represented as

           $P(X < 12) = P(\frac{X - \mu }{\sigma_{x}} < \frac{12 - 12.01 }{0.02917 } )$

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

      $P(X < 12) = P(Z < -0.3425 )$

From the z table  the area under the normal curve to the left corresponding to   -0.3425  is

      $P(X < 12) = P(Z < -0.3425 ) = 0.3660$

=>  $P(X < 12) = 0.3660$

Considering question a

From the question we are told that

   The mean is  $\mu = 12.03 \ oz$

Generally the  probability that the mean volume of a random sample of 144 bottles is less than 12 oz is mathematically represented as

           $P(X < 12) = P(\frac{X - \mu }{\sigma_{x}} < \frac{12 - 12.03 }{0.02917 } )$

          $P(X < 12) = P(Z < -1.0274 )$

From the z table  the area under the normal curve to the left corresponding to   -1.0274   is

      $P(X < 12) = P(Z < -1.0274 ) =0.1521$

=>   $P(X < 12) = 0.1521$