Normal curve

One way to analyze aset of data values is tograph its frequency curve. The Standard Normal CurveThis bell-shaped curve is the frequency curve of anormal distribution. The z-scores provide a standard measure where N(0, 1) representsthe standard normal curve. The area under the curve relates tothe probability of an event occurring with the total area equaling one.9.3.4Example: If the frequency of the lifetime of several thousand light bulbs were plotted, the graph wouldresemble a bell-shaped curve.

Normal curveThe curve is symmetric about the mean. Each half represents 50% of the total area.The total area is 1.0000Areas can be thought of as probabilities.Areas could be written as percents.Areas can not be negative.

Standard Normal curve

The mean is 0.

The standard deviation is 1.

Format of the tableThe table has 2 halves one for negative values of z and one for positive values of z.The left column tells the first decimal place for the z; the top row tells the second decimal place for z.We find the intersection of the row and column to find the area to the LEFT of z.

Special notesFor Z scores above 3.50, use the area of 0.9999

For Z scores below -3.50, use the area of 0.0001

The Z ScoreWhat we need is a standardized normal curve which can be used for any normally distributed variable. Such a curve is called the Standard Normal Curve.*

Reminder!

Areas can never be negative.

Z scores can be negative.

Area under the Standard Normal Distribution Curve1. To the left of any z value:Look up the z value in the table and use the area given.Bluman, Chapter 6*

Area under the Standard Normal Distribution Curve2. To the right of any z value:Look up the z value and subtract the area from 1.Bluman, Chapter 6*

Area under the Standard Normal Distribution Curve3. Between two z values:Look up both z values and subtract the corresponding areas.Bluman, Chapter 6*

Example 1Find the area to the left of z = 2.25

Example 2Find the area to the right of z = 1.50

Example 3Find the area between z = -1.35 and z = 2.15

Your TurnFind the area for each:1. area to the left of z = -1.04

2. area to the right of z = 1.07

Your turn continuedarea between z = 0 and z = 2.75

area between z = -1.00 and z = 1.00

Finding z given areaMake sure the diagram shows the area to the left of the desired z score.

Look in the body of the chart for the closest area, except for the special z scores.

Read the z score.

Example 4The area to the right of z is 0.0250

Example 5The area from the mean to a positive z is 0.1628

Example 6The area from some negative z to the mean is 0.4772

Your turnFind the z score for each: 5% is to the left of z

2. the top 15%

9.3.9Finding the Area Under the Curve1. Find the area between z-scores -1.22 and 1.44.-1.221.44-1.221.44The area for z-score -1.22 is0.1112.The area for z-score 1.44 is0.9251.Therefore, the area between z-scores -1.22 and 1.44 is0.9251 - 0.1112 = 0.8139.

Finding the Area Under the Curve2. Find the area between the mean and z-score -1.78. -1.78The area for z-score -1.78 is0.0375.Therefore, the area between the mean and z-score -1.78 is0.5 - 0.0375 = 0.4625.3. Find the area between the mean and z-score 1.78. 1.78The area for z-score 1.78 is0.9625.Therefore, the area between the mean and z-score 1.78 is0.9625 - 0.5 = 0.4625.9.3.10

Finding a Z-Score Given the Area1. Find the z-score for an area of 0.4901 that is left of the mean.2. Find the z-score for an area of 0.2177 greater than the z-score.3. Find the z-scores when approximately 80% of the data is evenly distributed about the mean.Find the area to the left of the z-score:0.5 - 0.4901 = 0.0099The z-score for an area of 0.0099 is -2.33.Find the area to the left of the z-score:1 - 0.2177 = 0.7823 The z-score for an area of 0.7823 is 0.78.Find the area to the left of the z-score:0.5 - 0.4 = 0.1The z-scores for 80% of the area are-1.29 and 1.29.9.3.12

Using Normal Distribution - Applications1. A television manufacturing company found that the life expectancy of their televisions was normally distributed with a mean of 15 000 h and a standard deviation of 1100 h.a) What percent of the televisions last longer than 17 000 h?b) What percent last between 12 600 h and 16 200 h?c ) In a shipment of 500 televisions, how many would you expect to last less than 13 000 h? a) Calculate the z-score for 17 000 h:1.82The area greater than z-score 1.82 is 1 - 0.9656 = 0.0344.Therefore, 3.44% of televisions last longer than 17 000 h.9.3.13

b) What percent last between 12 600 h and 16 200 h?Find the z-scores for 12 600 h and 16 200 h:1.09-2.18The area for z-score -2.18 is 0.0146.The area for z-score 1.09 is 0.8621.Therefore, the area between z-scores -2.18 and 1.09 is 0.8621 - 0.0146 = 0.8475.Therefore, 84.75% of the televisions would last between 12 600 h and 16 200 h.9.3.14Using Normal Distribution - Applications [contd]

In a shipment of 500 televisions, how many would you expect to last less than 13 000 h?Find the z-scores for 13 000 h:-1.82The area for z-score -1.82 is 0.0344.Therefore, 3.44% of the televisions would last less than 13 000.Therefore, you would expect 500 x 0.0344 = 17 televisions to last less than 13 000 h.9.3.15Using Normal Distribution - Applications [contd]

9.3.17Using Normal Distribution - Applications A staple machine holds 400 staples when loaded. The mean number of misfires per load is 8 with a standard deviation of 2.8. Assuming normal distribution, what is the probability that there will be fewer than 11 but more than 7 misfires per load?-0.361.07Area = 0.3594Area = 0.8577P(-0.36 z 1.07)= 0.8577 - 0.3594= 0.4983Therefore, the probability that there will be fewer than 11 but more than 7 misfires per load is 0.4983.

9.3.18Using Normal Distribution - Applications 3. A brush manufacturer determines the mean life of his brushes to be five years with a standard deviation of two years. If he guarantees his brushes for three years, what percent of his brushes will he have to replace?-1Area = 0.1587Therefore, 15.87% of the brusheswill have to be replaced.