mean 4.54, stand. dev. 0.25: percent > 5, between 4.4 & 4.6

Godisgood

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The lengths of the sardines received by a cannery have a mean of 4.54 inches and a standard deviation of 0.25 inch. If the distribution of these lengths has roughly the shape of a normal distribution, what percentage of all these sardines are: (a) longer than 5.00 inches; (b) from 4.40 to 4.60 inches long.
 
Re: I'm lost

I will start you off on this one and it should help you with similar ones. OK?.

part a: use \(\displaystyle z=\frac{x-{\mu}}{\sigma}\)

\(\displaystyle z=\frac{5-4.54}{.25}\)

Now, find this value and look it up in the z table. Since you want 'longer than 5', you want the region to the right. So, subtract it from 1.

The z values come from negative infinity and go to the right. The area under the entire normal curve is 1, so you subtract your value from 1 in order to get the area to the right. Because it gives you the area to the left of x=5.

For part b, use the same formula, look them up in the table and subtract them.
 
Re: I'm lost

Godisgood said:
The lengths of the sardines received by a cannery have a mean of 4.54 inches and a standard deviation of 0.25 inch. If the distribution of these lengths has roughly the shape of a normal distribution, what percentage of all these sardines are: (a) longer than 5.00 inches; (b) from 4.40 to 4.60 inches long.

You have posted 5 problems - without showing a line of work.

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
 
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