Statistics Problem, again!

[NPWagner]

A group of Florida State University psychologists examined the efects of alcohol on the reactions of people to a threat. after obtaining a specified blood alcohol level, experimental subjects were placed in a room and threatened with electric shocks. using sophisticated equipment to monitor the subjects' eye movements, the startle response (measured in milliseconds) was recorded for each subject. The mean and standard deviation of the startle were 37.9 and 12.4 respectively. Assume that the startle response x for a person with the specified blood alcohol level is approximately normaly distributed.

A. Find the probability that x is between 40 and 50 milliseconds.
B. Find the probability that x is less than 30 milliseconds.
C. Give an interval for x, centered around 37.9 milliseconds, so that the probability that x falls in the interval is .95.

Thanks for the help again!

 

[tkhunny]

the mean and standard deviation of the startle were 37.9 and 12.4 respectively.

A. Find the probability that x is between 40 and 50 milliseconds.
B. Find the probability that x is less than 30 milliseconds.
C. Give an interval for x, centered around 37.9 milliseconds, so that the probability that x falls in the interval is .95.

That's funny. My first thought was, "I expect they had no trouble finding volunteers at Florida State!" :lol:

You need Z-scores. I'll do the first one.

(40-37.9)/12.4 = 0.1694
(50-37.9)/12.4 = 0.9758

Then you ask youself (or a table or a calculator), how much Normal probability is between 0.1694 and 0.9758 standard deviations above the mean.

I have a table that gives the area from the mean to the given Z-Score, so,

Looking up 0.1694, I get (after interpolation) 0.0673
Looking up 0.9758, I get (after interpolation) 0.3355

Then I get 0.3355 - 0.0673 = 0.2682 or 26.82%

Using a fancy calculator, I get 26.815922%