Since
zscores are for the Standard Normal Distribution, 

Example (TI83): Find the probability that a zscore is between
1.5 and 2. That is, find P(1.5 ≤ z ≤ 2). 
Find probability or area 
Press 2^{nd} VARS [DISTR]. Scroll down to 2:normalcdf( Press ENTER. 

Enter 1.5,2) and press ENTER to get the answer .91044. The syntax is normalcdf(smaller z, larger z). 

Since
zscores are for the Standard Normal Distribution, 

Example (TI83): Find the zscore for an area of 0.25 to the left
of the zscore. 

Press 2^{nd} VARS [DISTR]. Scroll down to 3:invNorm Press ENTER. 

Enter .25) and press ENTER to get the answer .67. The syntax is invNorm(area to left of desired z). 
Since
the IQs are NOT Standard Normal Distribution, 

Example (TI83): Adult IQs are normally distributed with µ = 100
and σ = 15. Find the probability that a randomly selected IQ is less
than 112. That is, find P(x < 112).
Since this is a continuous function, we have P(x < 112) = P(x ≤ 112). 

Press 2^{nd} VARS [DISTR]. Scroll down to 2:normalcdf( Press ENTER. 

Enter 9999,112,100,15) The syntax is normalcdf(smaller, larger, µ, σ). Note:
The 9999 is used as the smaller
value to be at least 5 standard deviations from the mean. 

Find the probability that a randomly
selected IQ is at least 122. That is, find
P(x ≥ 122). 

Press 2^{nd} VARS [DISTR]. Scroll down to 2:normalcdf( Press ENTER. 

Enter 122,9999,100,15) The syntax is normalcdf(smaller, larger, µ, σ). Note:
The 9999 is used as the larger
value to be at least 5 standard deviations from the mean. 

Find the probability that a randomly
selected IQ is between 112 and 122. 

Press 2^{nd} VARS [DISTR]. Scroll down to 2:normalcdf( Press ENTER. 

Enter 112,122,100,15) and press ENTER The syntax is normalcdf(smaller, larger, µ, σ). 