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 and is needed to keep the Shuttle in orbit. Below 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? P(17500 < x < 17660) = Normalcdf(-2.286, 2.286) a.) b.) c.) mph P(x < 17500) = Normalcdf(-E99, ) InvNorm(.99) = = (X – 17580)/35
WARM – UP P(x<#) 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) = English: P(x < 90) = P(z < 0.875) = Normalcdf(-E99, 0.875) Jake did better than 80.9% of his peers. P(z < 0.75) = Normalcdf(-E99, 0.75) Jake did better than 77.3% of his peers.
Normal Proportion Calculations ●State the Problems as an Inequality: 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. ●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. Finding Values from Proportions
1.A recent Statistics Test produced a. What proportion of the class scored below a 75 on the test? P( x < 75 ) = P(z < ) normalcdf(-E99, ) = First, Convert 75 to a z-score Normcdf = Proportion or % InvNorm = z-score
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 90 th Percentile. b) Within the z-score formula substitute the z, the μ, the σ, and then solve for x. X = 93.7 = 94% InvNorm(.90) = % 90%
3.Between what scores does a student have to get to be in the Middle 50% of the class? < X < InvNorm(.25) = Q1Q1 Q3Q3 InvNorm(.75) = %
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 90 th Percentile. Then find the z-score for the 20 th Percentile b) Within two z-score formulas substitute the z and x and then solve for the μ and the σ. InvNorm(.90) = InvNorm(.20) =
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? X: Z: % %
1.If the GPA of seniors follows, what proportion of the senior class has a GPA below 3.0? P( x < 3.0 ) = P(z < ) normalcdf(-E99, ) =
2. If the GPA of seniors follows, what GPA would represent the top 5% of the senior class? X = InvNorm(.95) = % Hint: a) First find the z-score for the 95 th Percentile. b) Within the z-score formula substitute the z 95, the μ, and the σ, and then solve for x
Between what GPA’s does a student have to get to be in the Middle 50% of the senior class if the distribution is ? < X < InvNorm(.25) = Q1Q1 Q3Q3 50% 25%25% InvNorm(.75) = IQR = Q 3 – Q 1
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). 1) What proportion traveled more than 300 miles? 2) What proportion traveled more than 450 miles? 3) What proportion traveled less than 100 miles? 4) What distance, in miles, represents the 90 th Percentile? 5) What distance, in miles, do the top 5% represent? 6) What percent of students stay close to home by traveling within 5 to 60 miles? miles 355 miles 3.42% P(x > 300) = P(z > ) Normalcdf(1.0227, E99) P(x > 450) = P(z > ) Normalcdf(2.7273, E99) P(x < 100) = P(z < -1.25) Normalcdf(-E99, -1.25) InvNorm(.90) z = = (X – 210)/88 InvNorm(.95) z = = (X – 210)/88 P(5 < x < 60) = P( < z < ) Normalcdf( , )