Question

You may need to use the appropriate appendix table or technology to answer this question. A certain organization reported the following mean scores for two parts of the Scholastic Aptitude Test (SAT). Evidence-based Reading and Writing | 533 Mathematics 527 Assume that the population standard deviation on each part of the test is o = 100. (Round your answers to four decimal places.) (a) What is the probability a sample of 66 test takers will provide a sample mean test score within 10 points of the population mean of 533 on the Evidence-based Reading and Writing part of the test? (b) What is the probability a sample of 66 test takers will provide a sample mean test score within 10 points of the population mean of 527 on the Mathematics part of the test? (c) Comment on the differences between the values computed in parts (a) and (b). The probability of being within 10 of the mean on the evidence-based reading and writing portion of the test is --Select-- probability of being within 10 on the Mathematics portion of the SAT. This is because the standard error is ---Select-- cases. the in both

Answer 1

a)

P(µ-10<x<µ+10) =P(523<x<543)

Z1 =   (X1 - µ )/(σ/√n) = ( -10 ) / (   100   / √   66   ) =   -0.81  
Z2 =   (X2 - µ )/(σ/√n) = ( 10 ) / (   100   / √   66   ) =   0.81  
                                          
P (   523   < X <    543   ) =    P (    -0.81   < Z <    0.81   )       
                                          
= P ( Z <    0.81   ) - P ( Z <   -0.81   ) =    0.792   -    0.208   =    0.5834       (answer)

b)

P(µ-10<x<µ+10)

Z1 =   (X1 - µ )/(σ/√n) = ( -10 ) / (   100   / √   66   ) =   -0.81  
Z2 =   (X2 - µ )/(σ/√n) = ( 10 ) / (   100   / √   66   ) =   0.81  
                                          
P (   523   < X <    543   ) =    P (    -0.81   < Z <    0.81   )       
                                          
= P ( Z <    0.81   ) - P ( Z <   -0.81   ) =    0.792   -    0.208   =    0.5834       (answer)

c)

same as

same