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

Respuesta :

Answer:

The value is [tex]x = 5.0744 \ ounce[/tex]

Step-by-step explanation:

From the question we are told that

  The mean is  [tex]\mu  = 5 \ ounce[/tex]

  The standard deviation is [tex]\sigma =  0.3 \ ounce[/tex]

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

    [tex]P(X  > x ) =P( \frac{X - \mu}{\sigma } >  \frac{x -5}{0.3}  ) =  0.098[/tex]

=> [tex] P(X  > x ) =P( Z >  \frac{x -5}{0.3}  ) =  0.098[/tex]

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

   [tex]z =0.248[/tex]

So

     [tex]\frac{x -5}{0.3} = 0.248[/tex]

=>   [tex]x = 5.0744 \ ounce[/tex]