A particular fruit’s weights are normally distributed, with a mean of 353 grams and a standard deviation of 6 grams. If you pick one fruit a

Question

A particular fruit’s weights are normally distributed, with a mean of 353 grams and a standard deviation of 6 grams. If you pick one fruit at random, what is the probability that it will weigh between 334 grams and 344 grams?

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Adalynn 2 weeks 2021-09-09T11:39:25+00:00 1 Answer 0

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    2021-09-09T11:41:11+00:00

    Answer:   0.0660

    Step-by-step explanation:

    Given : A particular fruit’s weights are normally distributed with

    Mean : \mu=353\text{ grams}

    Standard deviation : \sigma=6\text{ grams}

    The formula to calculate the z-score is given by :-

    z=\dfrac{x-\mu}{\sigma}

    Let x be the weight of randomly selected fruit.

    Then for x = 334 , we have

    z=\dfrac{334-353}{6}=-3.17

    for x = 344 , we have

    z=\dfrac{344-353}{6}=-1.5

    The p-value : P(334<x<353)=P(-3.17<z<-1.5)

    P(-1.5)-P(-3.17)=0.0668072-0.000771=0.0660362\approx0.0660

    Thus, the probability that it will weigh between 334 grams and 344 grams = 0.0660.

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