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4 years ago

14

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?

1 answer:

iVinArrow [24]4 years ago

3 0

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 : $P(334$

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

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

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Answer:

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Using <em>factorization</em> method:

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Taking $2y^{2}$ common:

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