document.write( "Question 1197365: The number of chocolate chips in an 18-ounce bag of a particular brand of chocolate chip cookies is normally distributed with mean 1,173 chips and standard deviation 115 chips. \r
\n" ); document.write( "\n" ); document.write( "Calculate the 30th percentile of the number of chocolate chips in an 18-ounce bag of these chocolate chip cookies. Give your answer as a decimal rounded to two digits after the decimal.
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Algebra.Com's Answer #830627 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "mu = 1173 = population mean
\n" ); document.write( "sigma = 115 = population standard deviation\r
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\n" ); document.write( "\n" ); document.write( "On a TI83 or TI84 calculator, you will type in the following
\n" ); document.write( "invNorm(0.3)
\n" ); document.write( "This function is found by pressing the key labelled \"2nd\" followed by the VARS key. The invNorm function is the third item.
\n" ); document.write( "The result is approximately -0.5244
\n" ); document.write( "Here's another calculator you can use
\n" ); document.write( "https://davidmlane.com/normal.html
\n" ); document.write( "Alternatively, you can use the NORMINV function in a spreadsheet.\r
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\n" ); document.write( "\n" ); document.write( "What this tells us is
\n" ); document.write( "P(Z < -0.5244) = 0.30 approximately
\n" ); document.write( "In other words, z = -0.5244 is the approximate 30th percentile critical z score.
\n" ); document.write( "About 30% of the area under the Z curve is to the left of z = -0.5244\r
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\n" ); document.write( "\n" ); document.write( "Use this z score to find the corresponding x value
\n" ); document.write( "z = (x - mu)/sigma
\n" ); document.write( "z*sigma = x-mu
\n" ); document.write( "x = z*sigma + mu
\n" ); document.write( "x = -0.5244*115 + 1173
\n" ); document.write( "x = 1112.694
\n" ); document.write( "x = 1112.69\r
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\n" ); document.write( "\n" ); document.write( "Answer: 1112.69
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