1 Answer

Mohammad Anini · Answer 1 · 2021-12-25T12:01:46+0000

Answer:

a) About 95% of organs will be within 230 and 430 grams.

b) 99.7%

c) 0.3%

d) 83.85%

Explanation:

The empirical rule 68-95-99.7 tells us that:

We have a bell shaped distribution (we can assume as approximately normal) with mean of 330 g. and standard deviation of 50 g.

a) This happens for an interval with ±2 standard deviations from the mean.

That is:

$X_1=\mu+z_1\cdot\sigma=330-2\cdot 50=330-100=230\\\\ X_2=\mu+z_2\cdot\sigma=330+2\cdot 50=330+100=430$

About 95% of organs will be within 230 and 430 grams.

b) We can calculate the z-scores for each value to know how many standard deviations are from the mean.

$z_1=(X_1-\mu)/(\sigma)=(180-330)/(50)=(-150)/(50)=-3\\\\\\z_2=(X_2-\mu)/(\sigma)=(480-330)/(50)=(150)/(50)=3\\\\\\$

As the values are 3 standard deviations from the mean each, it is expected that 99.7% of the organs weigh between 180 and 480 grams.

c) This is the complementary of the point b.

Then, it is expected that (100-99.7)%=0.3% of the organs weigh less than 180 grams or more than 480 grams.

d) We can calculate the z-scores for each value to know how many standard deviations are from the mean.

$z_1=(X_1-\mu)/(\sigma)=(280-330)/(50)=(-50)/(50)=-1\\\\\\z_2=(X_2-\mu)/(\sigma)=(480-330)/(50)=(150)/(50)=3\\\\\\$

Between 280 and 330 there is 68%/2=34% of the data.

Between 330 and 480 there is 99.7%/2=49.85% of the data.

Then, between 280 grams and 480 grams there is (34+49.85)%=83.85% of the data.

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