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prob of failing

", "templateType": "anything", "group": "Ungrouped variables", "definition": "normalcdf(pass,0,1)"}, "pass_rate": {"name": "pass_rate", "description": "

passrate

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A large group of students took a test in Maths and the final grades have a mean of \\(\\var{m1}\\)% and a standard deviation of \\(\\var{sd}\\)%. 

", "metadata": {"licence": "Creative Commons Attribution-NonCommercial 4.0 International", "description": ""}, "ungrouped_variables": ["m1", "sd", "n", "m2", "p1", "pass", "pfail", "pass_rate"], "parts": [{"showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "marks": 0, "type": "gapfill", "gaps": [{"showFeedbackIcon": true, "minValue": "(1-{p1})*100", "variableReplacements": [], "correctAnswerStyle": "plain", "precisionType": "dp", "allowFractions": false, "precision": "1", "correctAnswerFraction": false, "precisionMessage": "You have not given your answer to the correct precision.", "mustBeReduced": false, "mustBeReducedPC": 0, "scripts": {}, "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "maxValue": "(1-{p1})*100", "marks": 1, "precisionPartialCredit": 0, "type": "numberentry", "showPrecisionHint": false, "strictPrecision": false, "notationStyles": ["plain", "en", "si-en"]}], "showCorrectAnswer": true, "prompt": "

If we approximate the distribution of these grades by a normal distribution, what percentage of the students scored higher than \\(\\var{m2}\\)%? 

\n

Give your answer correct to one decimal place.      [[0]]%

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If the pass mark is \\(\\var{pass_rate}\\)% what percentage of the students should fail the test?  

\n

Give your answer correct to one decimal place.       [[0]]%

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\\(X\\) has a mean of \\(\\var{m1}\\)% and a standard deviation of \\(\\var{sd}\\)%. 

\n

(a)    Find the percentage of the students scored higher than \\(\\var{m2}\\)%

\n

\\(P(X>\\var{m2})=P\\left(Z>\\frac{\\var{m2}-\\var{m1}}{\\var{sd}}\\right)\\)

\n

\\(=P(Z>\\var{n})\\)

\n

\\(=1-P(Z<\\var{n})\\)

\n

\\(=1-\\var{p1}\\)

\n

\\(=\\simplify{1-{p1}}\\)

\n

The percentage of the students scored higher than \\(\\var{m2}\\)% in the exam  =  \\(\\simplify{100*(1-{p1})}\\)%

\n

\n

(b)   \\(P(Failing)=P(X<\\var{pass_rate})\\)

\n

  \\(=P\\left(Z<\\frac{\\var{pass_rate}-\\var{m1}}{\\var{sd}}\\right)\\)

\n

  \\(=P(Z<\\var{pass})\\)

\n

  \\(=1-P(Z<\\simplify{-{pass}})\\)

\n

  \\(=1-\\simplify{1-{pfail}}\\)

\n

  \\(=\\var{pfail}\\)

\n

The percentage of the students that should fail the test is thus \\(\\simplify{100*{pfail}}\\)%

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