Scores on an examination are assumed to be normally

Step 1 of 8

Let y denote the scores on an examination. Given that y follows a normal distribution with a mean of 78 and variance of 36.

(a) 

We want to determine the probability of a person taking the examination scores higher than 72. That is

\(P\left( {y \ge 72} \right) = P\left( {\frac{{y - \mu }}{\sigma } \ge \frac{{72 - \mu }}{\sigma }} \right)\)

\(P\left( {y \ge 72} \right) = P\left( {z \ge \frac{{72 - 78}}{6}} \right)\)

\(P\left( {y \ge 72} \right) = P\left( {z \ge  - 1} \right)\)

\(P\left( {y \ge 72} \right) = 1 - 0.1587\)

\(P\left( {y \ge 72} \right) = 0.8413\)

Hence, the probability that a person taking the examination scores higher than 72 is 0.8413

Step 2 of 8

b) 

Now, we want to determine a student's minimum marks for receiving an A grade.

Suppose the student gets x marks, then he receives an A grade.

That is 

\(P\left( {y \ge x} \right) = 0.1\)

\(P\left( {\frac{{y - \mu }}{\sigma } \ge \frac{{x - \mu }}{\sigma }} \right) = 0.1\)

\(1 - \phi \left( {\frac{{x - 78}}{6}} \right) = 0.1\)

\(\frac{{x - 78}}{6} = 1.282\)

\(x = 85.692\)

Therefore, the student gets a minimum of 85.692 marks and receives an A grade.

Step 3 of 8

c) 

The cut-off point for passing the examination is if the examiner wants only the top 28.1% of all scores to be passed.

That is

\(P\left( {y \ge {y_0}} \right) = 0.281\)

\(P\left( {\frac{{y - \mu }}{\sigma } \ge \frac{{{y_0} - \mu }}{\sigma }} \right) = 0.281\)

\(1 - \phi \left( {\frac{{{y_0} - 78}}{6}} \right) = 0.281\)

\(\frac{{{y_0} - 78}}{6} = 0.5798\)