A particular fruit's weights are normally distributed, with a mean of 651 grams and a standard deviation of 27 grams. if you pick 49 fruit at random, what is the probability that their mean weight will be between 642 grams and 657 grams

Respuesta :

Answer: About 21.64% of the fruit will be between 642 and 657 grams.

To solve this problem, we will have to find the probability for each z-score on the normal distribution. Then, subtract them to find the difference.

Lower End:
-9/27 = -0.33 for the z-score which is a probability of 0.3707

Upper End: 
6/27 = 0.22 for the z-score which is a probability of 0.5871.

0.5871 - 0.3707 = 0.2164

Or about 21.64%

The probability of mean weight [tex]642[/tex] gram is [tex]p=0.0990[/tex]

The probability of mean weight [tex]657[/tex] gram is  [tex]p=0.9946[/tex]

Given Equation:

    Mean [tex]=651[/tex]

  Standard deviation [tex]=27[/tex]

   n[tex]=49[/tex]

    The general form of,

   

           [tex]z=\frac{x -\mu}{\sigma\ /\sqrt{n} }[/tex]

Probability of [tex]657[/tex]:

By given,

 

            [tex]x=642\\\sigma=27\\n=49\\\mu = 651\\[/tex]

Substitute the values in the general form,

 

             [tex]z=\frac{642-651}{27/\sqrt{49} } \\\\z=\frac{-9}{27/7} } \\\\z=\frac{-9}{3.86} \\\\z=-2.331[/tex]

             [tex]p=0.0990[/tex]

Probability of [tex]657[/tex]:

 By given,

           [tex]x=657\\\sigma=27\\n=49\\\mu = 651\\[/tex]

Substitute the values in the general form,

            [tex]z=\frac{657-651}{27/\sqrt{49} } \\\\z=\frac{6}{27/7} } \\\\z=\frac{6}{3.86} \\\\z=1.55[/tex]

            [tex]p=0.9946[/tex]

For more information,

https://brainly.com/question/9334808