What % of Seniors only apply to 3 or less colleges? WARM UP The distribution for the number of colleges a High School Senior applies to is approximately normal with Mean 6 and Standard Deviation of 1. What % of Seniors only apply to 3 or less colleges? What % of Seniors apply to between 5 or 7 colleges? How many colleges do the top 2.5% of Seniors apply to? (100% - 99.7%)/2 = 0.15% 68% 8 or more 68% 3 4 5 6 7 8 9 3 σ from Mean = 99.7% 2 σ from Mean = 95%

Do NOT write down Method #1. Book - TABLE Z Chapter 6 (cont.): To CONVERT z-scores to Proportion or Area under a Standard Normal Curve there are Three Methods: Use Calculus: Integrate the following function from the Lower bound, a, to the Upper bound, b: Do NOT write down Method #1. Book - TABLE Z Green Chart - Table A Use the Calculator: 2nd VARS: #2 normalcdf( normalcdf(“Lower bound = a”, “Upper bound = b”) P(z < b) = normalcdf(-E99, b) P(z > a) = normalcdf(a, E99) P(a < z < b) = normalcdf(a, b)

P(-1 < z < 1) = normalcdf(-1, 1) P(-2 < z < 2) = = 0.683 normalcdf(-2, 2) = 0.954 normalcdf(-E99, 0) = 0.500

Use the Calculator: 2nd VARS: #2 normalcdf( normalcdf(“lower bound = a”, “Upper bound = b”) P(z < b) = normalcdf(-E99, b) P(z > a) = normalcdf(a, E99) P(a < z < b) = normalcdf(a, b) 0.9429 EXAMPLE: P( z < 1.58) = P(z < 1.33) = 0.9082

EXAMPLE: 1. P( z > 2.68) = 2. P(-1.02 < z < 1.33) = 0.0037 0.7544

P(z > - 0.75) and P(z < 0.75) = P(-0.75 < z < 0.75) Solve: |z| < 2 z > -2 z < 2 3. P(|z| < 0.75) = P(z > - 0.75) and P(z < 0.75) = P(-0.75 < z < 0.75) 0.5467 4. P(|z| > 0.75) = P(z < - 0.75) or P(z > 0.75) 0.4533

1. What z-score represents the TOP 15%? To CONVERT Proportions (%) to z-scores use the Calculator: 2nd VARS: #3 InvNorm(0.xx) This gives you the z-score that has xx% of the area to the Left. 1. What z-score represents the TOP 15%? 2. What z-score represents the Lowest 10%? 3. What z-scores represent the Middle 98%? 1.036 -1.282 -2.326, 2.326 98% 1% 1% 85% 15% 10%

Normcdf(za,zb) = Proportion InvNorm(0.xx) = z - score Normcdf(za,zb) = Proportion 1. A recent Statistics Test produced a . What proportion of the class scored below a 75 on the test? 68 74 80 86 92 98 104 -3 -2 -1 0 1 2 3 First, Convert 75 to a z-score P( x < 75 ) = P(z < -1.833) normalcdf(-E99, -1.833) = 0.0334

DO HW on a SEPARATE SHEET OF PAPER HW: Page 126: 29-34 DO HW on a SEPARATE SHEET OF PAPER

HW: Page 126: 29-34

HW: Page 126: 29-34

2. A recent Statistics Test produced a 2. A recent Statistics Test produced a . What score did a statistics student need to have earned in order to be within the Top 10% of the class? Hint: a) First find the z-score for the 90th Percentile. b) Within the z-score formula substitute the z, the μ, the σ, and then solve for x. 90% 10% InvNorm(.90) = 1.2816 X = 93.7 = 94%

3. Between what scores does a student have to get 3. Between what scores does a student have to get to be in the Middle 50% of the class? InvNorm(.25) = -0.674 InvNorm(.75) = 0.674 Q1 Q3 81.956 < X < 90.044

4. The ACT exam scores follow a normal distribution 4. The ACT exam scores follow a normal distribution. If the top 10% scored above a 34 and the bottom 20% scored below a 18, what is the mean and standard deviation? Hint: a) First find the z-score for the 90th Percentile. Then find the z-score for the 20th Percentile b) Within two z-score formulas substitute the z and x and then solve for the μ and the σ. InvNorm(.90) = 1.282 InvNorm(.20) = -0.842

Normal Proportion Calculations ● State the Problems as an Inequality: P(x < #) or P(x > #) or P(a < x < b) ● Draw a picture if necessary. ● Standardize the value by using the z-score formula. ● Use the Table or the Calculator [normalcdf (lower, upper)] to find the proportion. Finding Values from Proportions ● Draw a picture if necessary. ● Find the Standardized value (z) by using [InvNorm(.%)]. ● Solve for the x within the z-score formula.

The Math SAT has a Normal Distribution, If a student scores a 740, what would his z-score be? What Proportion of students score above 740? P( x > 740 ) = P(z > 2.53) = normalcdf(2.53, E99) = 0.0057

Upon reaching orbit the Space Shuttle will engage in MECO (Main Engine Cut-Off). At that time it will be traveling at an average of 17,580mph (σ = 35mph). A speed between 17500 and 17660 is needed to keep the Shuttle in orbit. Below 17500 is too slow. a.) What proportion of Shuttles reach this target interval? b.) What proportion of Shuttles have to fire secondary engines to correct speed because they are going to slow? c.) The top 1% of Shuttles accelerate over what speed? a.) 0.978 b.) 0.011 c.) 17661.4 mph P(17500 < x < 17660) = Normalcdf(-2.286, 2.286) P(x < 17500) = Normalcdf(-E99, -2.286) InvNorm(.99) = 2.326 2.326 = (X – 17580)/35