The heaviest 10% of fruits weigh more than 814.79 grams.
The problem setup implies that the weight of this fruit is a random variable when it says a particular fruit's weights are normally distributed.
Define X = Random Variable of the fruit's weight
[tex]X \sim N(mean = 760, standard deviation = 39)[/tex]
We want to find the specific gram value for which 8% of the fruits are heavier
The 92nd percentile of X's normal distribution
[tex]Solve for x: Pr( X < x ) = Normal( 760, 39 ) CDF = .92[/tex]
Standardizing we obtain
[tex]Pr( [X - 760]/39 < [x - 760]/39 ) = Pr( Z < [x - 760]/39 ) = .92,[/tex]
where Z is a standard normal random variable
[tex]Pr( Z < [x - 760]/39 ) = Standard \ Normal \ CDF( [x - 760]/39 )[/tex]
==> Set this equal to .92:
[tex]Standard \ Normal \ CDF( [x - 760]/39 ) = .92[/tex]
Then, from the standard normal table:
[tex][x - 760]/39 = Inverse Standard Normal[/tex]
[tex]CDF( .92 ) = 1.405[/tex]
Use algebra to solve for x:
[tex]x = 39 * 1.405 + 760 = 814.795[/tex]
Therefore the heaviest 10% of fruits weigh more than 814.79 grams.
To learn more about the weights normally distributed visit:
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