SOLUTION: The marks obtained by students in a mathematics test had an unknown mean and standard deviation. It is known that 12% of students gained a mark greater than 90, and 20% of the stu

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Question 1081613: The marks obtained by students in a mathematics test had an unknown mean and standard deviation. It is known that 12% of students gained a mark greater than 90, and 20% of the students scored a mark less than 60. Determine the value of mean and standard deviation assuming that the data is normally distributed.
Answer by stanbon(75887) (Show Source):
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The marks obtained by students in a mathematics test had an unknown mean and standard deviation. It is known that 12% of students gained a mark greater than 90, and 20% of the students scored a mark less than 60. Determine the value of mean and standard deviation assuming that the data is normally distributed.
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Note:: The 12$ have a left-tail of 88%
Find the z-value with a left-tail of 0.88
invNorm(0.88) = 1.1749
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So:: 90 = 1.1749*s + u
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Note:: The 20% is a left-tail
Find the z-value with a left-tail of 20%
invNorm(0.20) = -0.8146
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So 60 = -0.8146*s + u
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Equation:: 90-1.1749s = 60+0.8146s
30 = 2.0165s
s = 14.88
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Solve for "u"::
u = 60 + 0.9146*14.88
u = 73.61
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Cheers,
Stan H.
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