Standardized score 116 10016 1 Your IQ is 1 standard deviation above the mean

# Standardized score 116 10016 1 your iq is 1 standard

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Standardized score = (116 100)/16 = + 1 Your IQ is 1 standard deviation above the mean. Suppose your IQ score was 84. Standardized score = (84 100)/16 = 1 Your IQ is 1 standard deviation below the mean. Chapter 8 - Continuous Probability Distributions 01/03/2015 66
Example Compute the following probabilities (a) P(Z ≤ 1.25) = Φ( 1.25)=0.89435 (b) P( Z > 1.25)= 1 - Φ(1.25) = 0.10565 (c) P(Z - 1.25)= Φ( -1.25)= 0.1056 5 ( By symmetry of the normal curve , this is the same answer as (b). (d) P( - 0.38 Z 1.25 ) =Φ(1.25) - Φ( -0.38) =0.89435 - 0.35197 = 0.54238 Chapter 8 - Continuous Probability Distributions 01/03/2015 67
Percentiles and Standardized Scores Finding percentiles for normal curves requires: 1- Your own value . 2- The mean for the population of values . 3- The standard deviation for the population( s.d) . 4- Find the standardized score Note : keep the plus or minus sign. 5- Table can be used to find percentiles. Chapter 8 - Continuous Probability Distributions 01/03/2015 68
Percentile of the Standard Normal Distribution The 99 th percentiles of the standard normal distribution Is that value on the axis such that the area under the curve to the left of the value is 0.99. We have the area and we want the value of z. The Z table is used in inverse fashion. Chapter 8 - Continuous Probability Distributions z curve area=0.99 99th percentile 01/03/2015 69
Nonstandard Normal Distribution The lifetime of a battery is normally distributed with a mean life of 40 hours and a standard deviation of 1.2 hours. Find the probability that a randomly selected battery a) Lasts between 42 to 43 hours P( 42 x 43)= P((42-40)/1.2 z (43-40) / 1.2) P( 1.67 z 2.5) = Φ (2.5) - Φ (1.67) = 0.99379 0.95254 b) Lasts longer than 42 hours P( x>42)= 1- p(x ≤ 42) =1 - 0.95254= 0.04746 (approximately 4.8%) Chapter 8 - Continuous Probability Distributions 01/03/2015 70
Example Scores on the SAT Reading test in recent years follow approximately the N(504,111) distribution. How high a student score to place in the top 10% of all students taking SAT?? Note: N( μ, σ ) Chapter 8 - Continuous Probability Distributions 01/03/2015 71
1) State the problem and draw a picture Area=0.10 Chapter 8 - Continuous Probability Distributions Area=0.9 X= ? z=1.28 01/03/2015 72
The x- value that put the student in the top 10% is the same as the x- value for which 90% of the area is to the left of x look in the body of the Z-table for the entry closest to 0.9 It is 0.8997. This is the entry corresponding to z=1.2816( is the standardized value with area 0.9 to the left) Unstandardize Transform z back to the original x scale. z σ = x - μ x = z σ + μ x= mean+ (1.2816)(standard deviation) x= 504+(1.2816)(111)= 646.2576 Student most score at least 647 to place in the highest 10% Chapter 8 - Continuous Probability Distributions 01/03/2015 z = x - μ σ 73
Example If X is a normal random variable with parameters μ = 3 and σ 2 = 9, (a) P{2 < X < 5}= P{ (2-3) /3 < Z < (5- 3)/3}= Φ (2/3) – Φ( -1/3) = Φ(0.67 ) - Φ ( -0.33) = 0.74857 0.3707 = 0.37787 (b) P{X > 0} = 1- P{ X ≤ 0}= 1 - P{ Z ≤ (0 -3)/3) = 1- Φ( -1) = 1- 0.15866 = 0.84134 01/03/2015 Chapter 8 - Continuous Probability Distributions 74
( c) P{|X − 3| > 6} = P{X > 9} + P{X < −3} = P{Z > 2} + P{Z < −2} = 1 − Φ (2) + Φ(−2) = 0.0455 01/03/2015 Chapter 8 - Continuous Probability Distributions 75
The Empirical Rule: 68-95- 99.7” Rule If your data have a normal distribution , then approximately: 68% 76 Chapter 8 - Continuous Probability Distributions 01/03/2015
01/03/2015 Chapter 8 - Continuous Probability Distributions