• May 6th 2008, 05:07 AM
Hikaru
I can't figure this problem out. I left my maths book at school so I no longer have the formulas. :[

Q6. The mass of baby guinea pigs four days after birth is a normally distributed variable with a standard deviation of 2 grams.

a. What percentage of the guinea pigs are less than 4 grams below the mean weight?

b. What percentage of the guinea pigs are more than 6 grams above the mean weight?

I have my own graphics calculator too. I can get to a screen where you can enter these values...

Lower:
Upper:
Standard Deviation:
Mean:

I defiantly know we had to use that part of the calculator. If you guys can tell me how to work it out, that would be great!

• May 6th 2008, 05:16 AM
mr fantastic
Quote:

Originally Posted by Hikaru
I can't figure this problem out. I left my maths book at school so I no longer have the formulas. :[

Q6. The mass of baby guinea pigs four days after birth is a normally distributed variable with a standard deviation of 2 grams.

a. What percentage of the guinea pigs are less than 4 grams below the mean weight?

b. What percentage of the guinea pigs are more than 6 grams above the mean weight?

I have my own graphics calculator too. I can get to a screen where you can enter these values...

Lower:
Upper:
Standard Deviation:
Mean:

I defiantly know we had to use that part of the calculator. If you guys can tell me how to work it out, that would be great!

a. You're two standard deviations below the mean. So find Pr(Z < -2) and multiply the result by 100 to get percentage.

b. You're three standard deviations above the mean. So find Pr(Z > 3) and multiply the result by 100 to get percentage.

In both cases Z is a standard normal variable, mean = 0 and sd = 1.
• May 6th 2008, 05:40 AM
Hikaru
Quote:

Originally Posted by mr fantastic
a. You're two standard deviations below the mean. So find Pr(Z < -2) and multiply the result by 100 to get percentage.

b. You're three standard deviations above the mean. So find Pr(Z > 3) and multiply the result by 100 to get percentage.

In both cases Z is a standard normal variable, mean = 0 and sd = 1.

Ohhh that makes sense! =] I drew out a small graph/curvey thing and I get what you mean for the top half!

However I'm curious; Why exactly is the mean 0? And why is the Standard Deviation 1? In the question it says the Standard Deviation is 2...
• May 6th 2008, 06:06 AM
mr fantastic
Quote:

Originally Posted by Hikaru
Ohhh that makes sense! =] I drew out a small graph/curvey thing and I get what you mean for the top half!

However I'm curious; Why exactly is the mean 0? And why is the Standard Deviation 1? In the question it says the Standard Deviation is 2...

You should have learned about the standard normal distribution.

And you should have learned that $Z = \frac{X - \mu}{\sigma}$ where Z is standard normal random variable and X is a normal random variable.

Read this: Standard Normal Distribution (1 of 2)
• May 12th 2008, 08:03 AM
abender
Hi,

Heard of the 67-95-99.8 rule?

Given a normal distribution, 67 percent of the guinea pigs will be within 1 standard deviation of the mean. 95 percent will be within two standard deviations, and 99.8 are within three SD's. It is just a shortcut that is helpful to memorize, unless you like reading through those Z tables with tiny font(Rofl).

-Andy