*Note: this is a draft document- work in progress. What is here is correct but needs polish and citations.*

- The total area under the normal curve is equal to 1.
- The probability that a normal random variable
*X*equals any particular value is 0. - The probability that
*X*is greater than*a*equals the area under the normal curve bounded by*a*and plus infinity (as indicated by the*non-shaded*area in the figure below). - The probability that
*X*is less than*a*equals the area under the normal curve bounded by*a*and minus infinity (as indicated by the*shaded*area in the figure below).

Sample problems using Normal Distribution Calculator here

10. An average light bulb manufactured by the Acme Corporation lasts 300 days with a standard deviation of 50 days. Assuming that bulb life is normally distributed, what is the probability that an Acme light bulb will last at most 365 days? *Solution:* Given a mean score of 300 days and a standard deviation of 50 days, we want to find the cumulative probability that bulb life is less than or equal to 365 days. Thus, we know the following:

- The value of the normal random variable is 365 days.
- The mean is equal to 300 days.
- The standard deviation is equal to 50 days.

We enter these values into the Normal Distribution Calculator and compute the cumulative probability. The answer is: P( X < 365) = 0.90. Hence, there is a 90% chance that a light bulb will burn out within 365 days. 11. Suppose scores on an IQ test are normally distributed. If the test has a mean of 100 and a standard deviation of 10, what is the probability that a person who takes the test will score between 90 and 110? *Solution:* Here, we want to know the probability that the test score falls between 90 and 110. The “trick” to solving this problem is to realize the following:

P( 90 < *X* < 110 ) = P( X < 110 ) – P( X < 90 )

We use the Normal Distribution Calculator to compute both probabilities on the right side of the above equation.

- To compute P( X < 110 ), we enter the following inputs into the calculator: The value of the normal random variable is 110, the mean is 100, and the standard deviation is 10. We find that P( X < 110 ) is 0.84.
- To compute P( X < 90 ), we enter the following inputs into the calculator: The value of the normal random variable is 90, the mean is 100, and the standard deviation is 10. We find that P( X < 90 ) is 0.16.

We use these findings to compute our final answer as follows:

P( 90 < *X* < 110 ) = P( X < 110 ) – P( X < 90 ) P( 90 < *X* < 110 ) = 0.84 – 0.16 P( 90 < *X* < 110 ) = 0.68

Thus, about 68% of the test scores will fall between 90 and 110.