Algebra.Com's Answer #430755 by Positive_EV(69)![\"\"](\"/images/stars/stars-08.gif\")  
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For each section, you'll need to find a z-value or set of z-values, and use a table to find the appropriate probabilities.
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document.write( " , where X is the value, mu; is the mean, and sigma is the standard deviation.
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document.write( "1)  . If you look up Z = 1.88 (most Z-tables round to the nearest hundreth) on a Z-table such as http://lilt.ilstu.edu/dasacke/eco148/ztable.htm, you will get the value .9699. The value on the Z-table given is the probability that a random Z-value is less than the Z-value on the table. Thus, the probability the student scores less than 90 is .9699, and the probability he scores above 90 is 1 - .9699 = .0301.
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document.write( "2)  . The probability on the Z-table corresponding to Z = .63 is .7357. Again, since this is the probability that the value has a lower Z-score than .63, the probability the student scores above an 80 is 1 - .7357 = .2643. In order to find the probability of scoring between 80-90, you will need to subtract the probability the student scores above a 90 from the probability the student scores above an 80. This value is .2643 - .0301 = .2342.
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document.write( "30  . Reading the value of Z = -1.88 from the table gives the value .0301. Since the table gives the probability of a score lower than the Z-value, and this is the value you are looking for in this case, the probability of scoring less than 60 is read directly from the table and is equal to .0301.
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