Assume critical reading scores for a standardized test are distributed as N(580, 100). Complete parts (a) through (d) below. Assume critical reading scores for a standardized test are distributed as N(580, 100). Complete parts (a) through (d) below.a. Find the test score at the 75th percentile.The 75th percentile is.(Round to the nearest integer as needed.)b. Find the test score at the 25th percentile.The 25th percentile is.(Round to the nearest integer as needed.)c. Find the interquartile range for test scores.The interquartile range is(Round to the nearest integer as needed.)d. Is the interquartile range larger or smaller than the standard deviation?The interquartile range isthan the standard deviation.

Given that The test scores are normally distributed. Mean (μ) = 580 Standard deviation (σ) = 100 We know that z=(x-μ)σ Interquartile range = (Q3-Q1) = (75th percentile -25th percentile) 75th percentile score (z) = P(z<0.675) = 0.75 75th percentile z score = 0.675 25th percentile score (z) = P(z<-0.675) = 0.25 25th percentile z score = -0.675a) Test score at the 75th percentile z=0.675 ; (μ) = 580 ; (σ) = 100 0.675=(x-580)100 (x-580) =0.675 x 100 (x) =580+67.5 =647.5 Test score at 75th percentile (Q3) = 647.5 b) Test score at the 25th percentile z=-0.675 ; (μ) = 580 ; (σ) = 100 -0.675=(x-580)100 (x-580) =-0.675 x 100 (x) =580-67.5 =512.5 Test score at 25th percentile (Q1) = 512.5 c) Interquartile range (IQR) = (Q3-Q1) (Q3) = 647.5 ; (Q1) = 512.5 (IQR) = (647.5-512.5) = 135 (IQR) = 135 d) The interquartile range is larger than the standard deviation . [135>100]

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Assume critical reading scores for a standardized test are distributed as N(580, 100). Complete parts (a) through (d) belo .....

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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