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yanalaym [24]
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?
Mathematics
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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