## Inverse Normal Continuity Corrections

Often with normal distribution problems, probabilities will be given and the mean, standard deviation or particular values of the continuous random variable will need to be found. These are called inverse normal problems.

Sometimes the values of the random variable in a normal distribution are discrete and acontinuity correction has to be used.

### Inverse Normal Problems

In this type of problem, the given value of the probability is found in the body of the tables and the corresponding z-value is found. Then the other given values are substituted into the formula Example

The weights of apples from a tree are normally distributed with a mean of 450 g.

If 30% of the apples weigh over 475 g, find the standard deviation of the weights of the apples.

Step 1 Draw a diagram. Step 2 Look up the z-value corresponding to 0.5 − 0.3 = 0.2 in the body of the standard normal distribution tables. Step 3 Substitute into the z-formula The standard deviation of the weights of apples is 47.7g

### Continuity Corrections

The normal distribution is the probability distribution of a continuous random variable. Sometimes a normal distribution is used to model a discrete random variable. e.g. Scores in an exam or rounded measurements.

When this happens a continuity correction is used.

If a measurement is given to the nearest unit, it could actually be between half of the unit on either side of the given measurement.

e.g. A measurement of 58 kg could be in the range, greater or equal to 57.5 kg and less than 58.5 kg. A continuity correction would take this into account.

Example

 800 metre runners at age 12 have been found to record times which are normally distributed with a mean of 155 seconds with a standard deviation of 15 seconds. What is the probability that a randomly selected runner will record a time between 140 seconds and 160 seconds inclusive? If X is the random variable representing the 800 metre times, the probability required is P(140 <≤ X <≤ 160) With continuity corrections, this becomes P(139.5 ≤ X <160.5)  The probability that a runner runs between 140 and 160 seconds is 0.4924 