Standardizing a Normal Distribution into a Standard Normal Distribution

Review of the Normal Distribution 1.‘Bell-Shaped’ & Symmetrical 2.Mean, Median, Mode Are Equal 4. Random Variable Has Infinite Range Mean Median Mode X f(X)

Many Normal Distributions By varying the parameters  and , we obtain different normal distributions There are an infinite number of normal distributions

Finding Probabilities Probability is the area under the curve! c d X f(X)f(X)

Which Table to Use? An infinite number of normal distributions means an infinite number of tables to look up!

Normal Distribution Tables Normal distributions differ by mean & standard deviation. Each distribution would require its own table. That’s an infinite number! X f(X) Instead we use a procedure to “standardize” each distribution to a benchmark. That’s an one table only!

 6.4 Standardized Normal Distribution One distribution! Normal Distribution Z   = 0 z = 1 Z Standardized Normal Distribution X  X  X x  Z X x   

Steps to turn x variable into z score 1.) Draw sketch, label mean and any relevant scores, then shade the region representing desired probability. 2.) For relevant x score use to find equivalent z score. 3.) Refer to table to find area. You may have to add.5, subtract.5, subtract two areas, etc.

Z  Z = 0  Z = 1.12 Standardizing Example Let x = 6.2, Find z score. Normal Distribution Standardized Normal Distribution X  X = 5  X = Z X x x       

Z  Z = 0  Z = 1.12 Z Last Step: Obtaining the Probability Standardized Normal Probability Table (Portion) Probabilities

Note The z score for the mean of a normal distribution is always zero (since for a standard normal distribution the mean is zero).

Z  Z = 0  Z = Example: Let mean = 5 and s.d. = 10 Find Prob(3.8  X  5) Normal Distribution.0478 Standardized Normal Distribution X  X = 5  X = Z X x x       

0  Z = Z.21 Example: Let mean = 5 and s.d. = 10 Find Prob(2.9  X  7.1) Normal Distribution Standardized Normal Distribution Shaded Area Exaggerated 5  X = X Z X Z X x x x x              

Z  Z = 0  Z = 1.30 Example: Mean is 5 and s.d. = 10 Find Prob(X  8) Normal Distribution Standardized Normal Distribution Z X x x        X  X = 5  X = 10 8

Z  Z = 0  Z = 1 Example: Prob(X  11) Normal Distribution Standardized Normal Distribution Z X x x        X  X = 5  X =

 z = 0  Z = 1.30 Z.21 Example: Prob(7.1  X  8) Normal Distribution Standardized Normal Distribution Z X Z X x x x x                x = 5  X = X

Homework Page (28-36 even)

Answers to p (28 – 36 even) 28.) a b c. 1 d ) a b c ) a b. 0 c d ) a..1026b ) a b c d..6554

6-5 Applications of the Normal Distribution Example 1: According to Automotive Lease Guide, the Porsche 911 sports car is among the vehicles that hold their best value. A Porsche 911 is expected to command a price of $48, 125 after three years. Suppose the prices are normal distributed with (after three years) a mean price of $ 48, 125 and a standard deviation of $1600. Find the probability that a randomly selected three- year old Porsche 911 will sell for a price between $46,000 and $49,000.

Solution = 48, 125 and = Find P (46,000 < x < 49, 000)  Z X Z X x x x x            46,000 – 48,  49,000 – 48, Normal Distribution 48,125  X = 10 46,00049,000 X TOTAL:.6170

Example 2 A racing car is one of the many toys manufactured by Mack Corporation. The assembly times for the car is normally distributed with a = 55 minutes and = 4 minutes. The company closes at 5 pm. If one worker starts assemble at 4 pm, what is the probability that she will finish this job before the company closes for the day?  

Solution They are asking for us to find P (x<60). Because she has up to an hr. to finish assembly before close. 55  X = 4 60 X Z X x x        TOTAL:.8944

Example 3 The life span of a calculator is normally distributed with a = 54 months and = 8 months. The company guarantees a calculator within 36 months of purchase. About what percentage of calculators made by this company are expected to be replaced?  

Solution Find P( x < 36), since only calculators that last 36 months or less will be replaced. X 54  X = 8 36 Z X x x        =.0122

Homework Page ( even)

Answers to pages 284 – 285 (38 – 46 even) 38.) a..5205b ) a..0548b ) a..2646b ) a..0038b ).0475

Section 6.6. Determining the z values when an area under the normal distribution curve is known

Z Z  Z = 0  Z = 1.31 Example 1 : Finding Z Values given Area Standardized Normal Probability Table (Portion) What Is Z Given P(Z) =.1217?

Finding Z given Area Example 2 Find the value of z such that the area under the standard normal curve in the right tail is Z  = Look up in the body of the z table (We can chose.4949 or.4951) If we chose.4951, we have z = 2.58.

Z  Z = 0  Z = 1.31 X  X = 5  X = 10 ? Finding X Values for Known Probabilities (Given z,, and ). Normal DistributionStandardized Normal Distribution.1217 XZ     [ ] x  Formula:

Calculator Problem Finding X value. The life span of a calculator is normally distributed with a = 54 months and = 8 months. What should the warranty period be to replace a calculator if the company does not want to replace more than 1% of all calculators sold?  

Calculator Problem Solution We are to find the value of x such that the area to the left of x is 1 percent (.01). If we look up.4900 (actually.4901), we get z = The area is to the left so the z we want is z = X  = 54  X = 8.01 So this must be =.4900

Now that we have z = -2.33, Find x value. The Formula gives: XZ   X = 54 + [ * 8] = 54 – = Thus, the company should replace all the calculators that start to malfunction within months of the date of purchase so that they will not have to replace more than 1% of the calculators.

Example 2 SAT test Almost all high school students who intend to go to college take the SAT test. In 2002, the mean SAT score of all students was Debbie is planning to take the test soon. Suppose the SAT scores of all students who take this test with Debbie will have a normal distribution with a mean of 1020 and standard deviation of 153. What should her score be on this test so that only 10% of all examinees score higher than she does?

Solution We are asked to find the value of x such that the area under the curve to the right of x is 10%. x  = 1020  = Look up.4000 and you get z = Now use formula to get corresponding x value.

Since z = 1.28, Find x value. The Formula gives: XZ   X = [ 1.28 * 153] = = Thus, if Debbie scores 1216 on the SAT, only about 10% of the examinees are expected to score higher than she does.

Homework Page 290 (54 – 62 even)

Answers to page 290 (54 – 62 even) 54.) a..51b. -.75c. -.82d ) a. 1.96b c d ) a. 403b c. 694 d e. 409f ) x = ) a. x = 51.36b. x = 56.8

Chapter 6 Review Page 304 (1- 13 omit 8)

Answers to page 304 (1-13 omit 8) 1.) a 2.) a 3.) d 4.) b 5.) a 6.) c 7.) b

Continued… 9.) a..1875b..9304c d ) a b..61c d ) a..8293b..0436c d ) a b ) a. i).0407ii).2666 iii).3050 iv).3897v).7528 b c..7845