standard deviation

Question

Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. The cheerleaders for a local professional basketball team must be between 65.5 and 68.0 inches. If a woman is randomly selected, what is the probability that her height is between 65.5 and 68.0 inches?

Answer 1

Step 1: Calculate z scores for both given x values. Formula for z score: z = (x-µ)/σ

 

z1= (65.5 -63.6)/2.5 =.76

 

z2= (68.0 -63.6)/2.5 = 1.76

 

Step 2: Find the probability of getting a z score in between these two z scores. P (.76 < z <  1.76)

 

You can do this on the calculator with normalcdf(.76, 1.76)=.1844 or by using a z table, finding the associated values, and subtracting the probability of .76 from the probability of 1.76. 

 

To get normalcdf on the calculator, press 2nd, then the Vars button, and then option 2. 

 

So, to recap, the P (65.5 < height < 68.0)= .1844 or about 18.44%.