Thread: Standard Deviation

1)The average temperature in an american state was recorded as 98.7 F with a standard deviation of 0.7 F. If this is a normal model what percentage of the towns in this state were below 97 F?

Is this question simply a case of taking the Z-Score as 1.7 and reading off the table which I believe gives the answer of 0.446?

2) Species A of an animal weighs average of 50 pounds and has standard deviation of 3.3.

Species B average is 57.5 with standard deviation of 1.7.

A zoo has a species A weighing 45 pounds and B at 54 pounds. Explaining fully determine which is more underweight.

Could someone please direct me as to how i would go about finding this answer please.

Originally Posted by rooney

1)The average temperature in an american state was recorded as 98.7 F with a standard deviation of 0.7 F. If this is a normal model what percentage of the towns in this state were below 97 F?

Is this question simply a case of taking the Z-Score as 1.7 and reading off the table which I believe gives the answer of 0.446?

2) Species A of an animal weighs average of 50 pounds and has standard deviation of 3.3.

Species B average is 57.5 with standard deviation of 1.7.

A zoo has a species A weighing 45 pounds and B at 54 pounds. Explaining fully determine which is more underweight.

Could someone please direct me as to how i would go about finding this answer please.

$\displaystyle \displaystyle{Z = \frac{X - \mu}{\sigma}}$. You don't seem to have used this formula correctly in (1).

Compare the z-score of each species.

Originally Posted by rooney

1)The average temperature in an american state was recorded as 98.7 F with a standard deviation of 0.7 F. If this is a normal model what percentage of the towns in this state were below 97 F?.

You gota lurve elementary statistics questions where the question setter does not really understand statistics well enough to set questions that make any sense. Suppose the correlation scale for temperature is comparable to the size of a state?

I'm not insisting that the beginning student of statistics understand such subtleties, but I do insist that text book authors do.

CB

Originally Posted by mr fantastic

$\displaystyle \displaystyle{Z = \frac{X - \mu}{\sigma}}$. You don't seem to have used this formula correctly in (1).

Compare the z-score of each species.

Made a bit of a schoole boy error....

So for part 1 the equation I require is:

(97-98.7) / 0.7 = 2.43

Then read down the Normal Model table which gives a percentile of about 0.0075? Does that sound roughly correct?!

And does anyone have any idea on the second question??

Originally Posted by rooney

Made a bit of a schoole boy error....

So for part 1 the equation I require is:

(97-98.7) / 0.7 = 2.43

Then read down the Normal Model table which gives a percentile of about 0.0075? Does that sound roughly correct?!

And does anyone have any idea on the second question??

Did you read the second sentence of my reply?

Originally Posted by rooney

Then read down the Normal Model table which gives a percentile of about 0.0075? Does that sound roughly correct?!

Sounds good

Originally Posted by rooney

And does anyone have any idea on the second question??

First find the z-scores and hence the probabilities for each of these scenarios