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Test Comparison and Z-Score Analysis

Analyzing test scores in Math and English to determine which test was performed better. Converting scores to z-scores and calculating percentile.

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Test Comparison and Z-Score Analysis

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  1. WARM – UP Jake receives an 88% on a Math test which had a N(81, 8). That same day he takes an English test N(81, 12)and receives a 90%. Which test did he do better on with respect to the rest of the class? Support your answer by converting each score into a z-score and then calculating the percent of students he did better than for each test (percentile = P(x<#). Math: P(x < 88) = P(z < 0.875) = Normalcdf(-E99, 0.875) Jake did better than 80.9% of his peers. English: P(x < 90) = P(z < 0.75) = Normalcdf(-E99, 0.75) Jake did better than 77.3% of his peers.

  2. Normal Proportion Calculations ● State the Problem 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 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(.%)]. (% to the LEFT) ● Solve for the x within the z-score formula.

  3. 1. A recent Math Test produced a . What proportion of the class scored above a 90 on the test? First, Convert 90 to a z-score P( x > 90 ) = P(z > 1.143 ) 0.127 normalcdf(1.143, E99) =

  4. 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 = z X = 93.7 = 94%

  5. 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 50% 81.956 < X < 90.044 IQR = 90.044 – 81.956 = 8.088

  6. 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 = z InvNorm(.20) = -0.842 = z

  7. 4. continued… InvNorm(.90) = 1.282 InvNorm(.20) = -0.842 - ( )

  8. HW: PAGE 127: 36, 37, 41, 42, 43, 47

  9. HW: PAGE 127: 36, 37, 41, 42, 43, 47

  10. HW: PAGE 127: 36, 37, 41, 42, 43, 47

  11. 4. If the GPA of seniors follows , what proportion of the senior class has a GPA below 3.0? -3 -2 -1 0 1 2 3 P(z < 0.9811) P( x < 3.0 ) = 0.8367 normalcdf(-E99, 0.9811) =

  12. 5. If the GPA of seniors follows , what GPA would represent the top 5% of the senior class? Hint: a) First find the z-score for the 95th Percentile. b) Within the z-score formula substitute the z95, the μ, and the σ, and then solve for x. 0.05 90% -3 -2 -1 0 1 2 3 InvNorm(.95) = 1.6449 X = 3.352

  13. 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

  14. Application A headache medicine is effective if it demonstrates relief (a score of 5 or higher) for over 75% of the population. If patients are asked to describe the effects of a new drug by rating it on a scale from 1(NO Effect) to 10(Complete Relief), and their responses follow a N(5.8, 1.3): 1. What Proportion of the population indicated a relief score of 3 or below? 2. What score represents the top 75% of the population? 3. What Proportion of the population indicate a relief score 5 or higher? 0.0156 4.923 73.08% 75% X: 1.9 3.2 4.5 5.8 7.1 8.4 9.7 Z: -3 -2 -1 0 1 2 3

  15. Between what GPA’s does a student have to get to be in the Middle 50% of the senior class if the distribution is ? InvNorm(.25) = -0.674 InvNorm(.75) = 0.674 Q1 Q3 50% 25% 25% 2.123 < X < 2.837 IQR = Q3 – Q1

  16. You are interested in finding the proportions of high school students who travel various miles for college. An approximately normal distribution of students reveal a N(210, 88). • What proportion traveled more than 300 miles? • What proportion traveled more than 450 miles? • What proportion traveled less than 100 miles? • What distance, in miles, represents the 90th Percentile? • What distance, in miles, do the top 5% represent? • What percent of students stay close to home by traveling within 5 to 60 miles? 0.1532 P(x > 300) = P(z > 1.0227) Normalcdf(1.0227, E99) 0.00319 P(x > 450) = P(z > 2.7273) Normalcdf(2.7273, E99) 0.1056 P(x < 100) = P(z < -1.25) Normalcdf(-E99, -1.25) 322.8 miles InvNorm(.90) z = 1.2816 1.2816 = (X – 210)/88 InvNorm(.95) z = 1.6449 355 miles 1.6449 = (X – 210)/88 P(5 < x < 60) = P(-2.3295 < z < -1.7045) 3.42% Normalcdf(-2.3295, -1.7045)

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