Question

The weights of items produced by a company are normally distributed with a mean of 5 ounces and a standard deviation of 0.3 ounces. What is the minimum weight of the heaviest 9.8% of all items produced

Answer 1

Answer:

The value is $x = 5.0744 \ ounce$

Step-by-step explanation:

From the question we are told that

  The mean is  $\mu = 5 \ ounce$

  The standard deviation is $\sigma = 0.3 \ ounce$

  Generally the minimum weight of the heaviest 9.8% is mathematically represented as

    $P(X > x ) =P( \frac{X - \mu}{\sigma } > \frac{x -5}{0.3} ) = 0.098$

=> $P(X > x ) =P( Z > \frac{x -5}{0.3} ) = 0.098$

From the normal distribution table  the z-score  for 0.098 is  

   $z =0.248$

So

     $\frac{x -5}{0.3} = 0.248$

=>   $x = 5.0744 \ ounce$