Statistics Normal Distribution: Finding Values
Normal Distributions: Finding Values Example. In a survey of women in the US (ages 20 – 29), the mean height was 64 inches with a standard deviation of 2.75 inches.
a. What height represents the 95th percentile? First, we need the z score that represents the 95th percentile – the z score that corresponds to a probability or area of .9500 P95 = 1.645
Using the formula x zσ μ , we can find the height for the 95th percentile. x 1.6449 2.75 64
b. What height represents the first quartile? Q1 = P25
c. What is the Interquartile Range? IQR = Q3 – Q1
Remember that the IQR represents the range of the middle 50% of the data.
Example. A brand of automobile tire has a life expectancy that is normally distributed, with a mean life of 30,000 miles and a standard deviation of 2500 miles. You sell this brand of tire and you want to give a guarantee for free replacement of tires that don’t wear well. How should you word your guarantee if you are willing to replace approximately 10% of the tires you sell?
You are willing to replace the bottom 10%. Bottom 10% = P10
Here’s my ad… Here at Bowman’s Tire and Appliance Emporium, we will guarantee this brand of tire up to 26,800 miles. What a deal folks. Come on down.