Given any normal variable, x with mean =  and standard deviation =  , we define the z-score or standard score (the z value of the standard normal distribution) as:

So given any value of a normal random variable, x, and its related mean and standard deviation, then the x value can be converted to a standard normal variable, z by the formula above and be used to find probabilities for the normal random variable as we did with the standard normal random variable, z.

We will have lots of oppurtunities to use this formula.

1. Know how to compute the standard score, z for a normal random variable, x.

A normal variable is one whose values appears to have a symmetrical distribution as that of a normal distribution.
 

The z-score or z values of a normal variable, x corresponds to the number of standard deviations of about the mean

.

Example: Find the z-score or z value of a normal variable whose value is 45 if its mean is 50 and standard deviation is 5.

The value of the random variable is x = 45, and  ,

, which means that x = 45 is 1 standard deviation below the mean since the z-value is negative 1.

Example: Find the z-score or z value of a normal variable whose value is 650 if its mean is 500 and standard deviation is 100.

The value of the random variable is x = 650,   and  ,

, which means that x = 650 is 1 ½ standard deviations above the mean since the z-value is positive.

So z = 1.5.

2. Know how to determine the probability associated with a normal random variable, x using the z-score from the standard normal distribution.

Knowing the z-values of a normal random variable by using the conversion formula above, the probabilities for various values of x can be determined as in previous section of probabilities of standard normal random variable.

Example: For and example students mean test scores were 80 and the standard deviation was 15, what is the probability that a students gets:

(a) between 70 and 95 on the test?

(b) > 85 on the test?

Assume that the test score is normally distributed, then its mean = 80, and its standard deviation = 15.

(a) The associated z-score for the 70 and 95 are
 

For x =70,  , So z = -0.67 (since lookup table only goes to 2 decimal places). For 95 , So z = 1 The P[-0.67 < z < 1] = Pr[z=1] - Pr[z=-0.67] = 0.84134474 - 0.25142882 = 0.5899 So Pr[70 < x < 95] = 0.5899 or 59 of the students between 70 and 95.

(b) x > 85 on the test?
 

For x =85,  , So z = 0.33 (since lookup table only goes to 2 decimal places). The P[z > 0.33] = 1 - Pr[z = 0.33] =  1 - 0.62929995 = 0.3707 So Pr[ x > 85] = 0.3707 or 37.07% of the students score greater than 85..

3. Know how to find the z score from the probability of a normal random variable.

From the probability of a standard normal distribution one can find the associate z value or x value for the normal random variable.

Example: The average age of clients attending a rehabilitation center for the past 2 years is 35 and the standard deviation is 3. For this population find the following:

Assume x is the age of clients attending the rehab. center over the past 2 years, the mean of x = 35, and standard deviation = 3

(a) The 90th percentile, P₉₀

(b) The age of 30% of the population, P₃₀

(c) The percentile rank of the clients with an age of 40.

(a) We first express P₉₀ in terms of z, since P₉₀ represents the value age of the clients that is represented by < 90% of the population. ,

So P₉₀ = 38.84
 

Since the standard normal reference table gives values Pr[z>a], then Pr[z=?] = 0.90. From the table the corresponding z value associated with a probability of 0.90 is z=1.28. . , So solve for x we get: . , and . , That is 90% of the client age is less that x = 38.84

(b) The age of 30% of the population, P30 Pr[0< z < ?] = 0.30 The closest prob. in the standard normal table is 0.2980594 which corresponds to a z value of z=-0.53. So use the z-score formula to find corresponding x value: , So, x = 33.41 P30 = 33.41

(c) The percentile rank of the clients with an age of 40. Find P% +> Pr[x<40] = ? Remember percentile rank is the probability < that %. So find z-value associated with a value for x = 40 Use formula , The probability associated with a z-value of z = 1.67 is 0.9525 from the reference table. So the percent rank of x = 40 is about 95.25% or P95

Workshop Problem: For a normal population with mean 20 and standard deviation of 2.5. Find the following:

(a) The third quartile

(b) The 40% percentile.

(c) the percentile rank of the data value 21.5

(d) The percentile rank of the data value 15.5