B Links to Course Sldes O Lecture on Feb 16 by zora x cengageBrain Login or Rex
ID: 3162418 • Letter: B
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B Links to Course Sldes O Lecture on Feb 16 by zora x cengageBrain Login or Rex x IAMAT toal Homework mod x c awebassign net/web/student/Assignment 5679616 10. o Tss points I Previous Answers Poostala 4E specimens of a certain type is selected, and the compressive strength of each specimen is determined. The mean and standard devia found to be well approximated by a normal arve. ple of concrete s 600, and the sample histogram are calculated as )Approximately what percentage of the sample observations are between 2400 and 3 (Round the answer to the nearest whole number) Approximately what percentage ample observations are outside the interval mom 1800 to 42007 cRound the answer the nearest whole number. (e) What can be said about the approximate percentage or observations between 1800 and 2400 (Round the answer nearest whole number to the (d) Why would you not use chebyshev's Rule to answer the questions posed in parts (a) (c)? Chebyshev's Rule does not apply to values in these ranges. Chebyshev's Rule is not used because the histogram is well approximated by a normal curve. Chebyshev's Rule is the best way to solve this problem Chebyshev's Rule typically gives much larger values than are appropriate for a normal distribution. Chebyshev's Rule does not apply to the normal distribution. Save Progres -5 points Poostat54 EC50. The average reading speed of students completing a speed reading course is 500 words per minute wom If the standard deviation is so wpm, find the z score associated with each of the following reading speeds. (Round the answers to two decimal places.) 535 wpm Need Help? LnesiExplanation / Answer
Problem-1:
a)
= 7000
= 500
standardize x to z = (x - ) /
P( 6500 < x < 7500) = P[( 6500 - 7000) / 500 < Z < ( 7500 - 7000) / 500]
P( -1 < Z < 1) = 0.3413+0.3413 = 0.6826
68.26%
(From Normal probability table)
b)
= 7000
= 500
standardize x to z = (x - ) /
P( 6000 < x < 8000) = P[( 6000 - 7000) / 500 < Z < ( 8000 - 7000) / 500]
P( -2 < Z < 2) = 0.4772+0.4772=0.9544
95.44% are inside the interval
Therefore, 100-95.44 = 4.56% are outside of 6000 to 8000
(From Normal probability table)
c)
= 7000
= 500
standardize x to z = (x - ) /
P( 6000 < x < 6500) = P[( 6000 - 7000) / 500 < Z < ( 6500 - 7000) / 500]
P( -2 < Z < -1) = 0.1587 - 0.0228 = 0.1359
13.59%
(From Normal probability table)
Values are ben chaning, working on it
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