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A particular fruit's weights are normally distributed, with a mean of 615 grams and a standard deviation of 10 grams.

The heaviest 10% of fruits weigh more than how many grams?

Give your answer to the nearest 0.1 gram.

Respuesta :

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.

What are the formula 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:

https://brainly.com/question/17091450

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