the result of a particular examination are given in the table below passed with distinction 10% passed without distinction 60% failed 30% it is known that a candidate fails in the examination if he obtains less than 40 marks while he must obtain at least 75%marks in order to pass with distinction . determine the mean and standard deviation of distribution of marks .assuming these is to be nominal

We are given that,

$p(x\lt 40)=0.3$ and $p(x\gt75)=0.1$. These are the two probabilities we are going to use in order to find the mean and the standard deviation as follows.

Standardizing,

$p(Z\lt {40-\mu\over \sigma})=0.3......(1)$ and $p(Z\gt {75-\mu\over \sigma})=0.1.........(2)$

From equation (1),

$\phi({40-\mu\over \sigma})=0.3\implies {40-\mu\over \sigma}=Z_{0.3}$ where $Z_{0.3}$ is the table value associated with 0.3.

From the standard normal tables, $Z_{0.3}= -0.5244005$. Therefore,

${40-\mu\over \sigma}=-0.5244005\implies \mu=40+0.5244005\sigma.....(3)$

From equation (2),

We write this equation as,

$p(Z\lt {75-\mu\over \sigma})=0.9\implies \phi( {75-\mu\over \sigma})=0.9\implies {75-\mu\over \sigma}=Z_{0.9}$ where, $Z_{0.9}$ is the table value associated with 0.9. From the standard normal tables, $Z_{0.9}=1.281552$. Therefore, ${75-\mu\over \sigma}=1.281552\implies \mu=75-1.281552\sigma.......(4)$

Equation 3 and 4 are equal so, equating both we have.

$40+0.5244005\sigma=75-1.281552\sigma\implies 1.8059525\sigma=35\implies \sigma=19.38(2dp)$

We can use either equation (3) or (4) to solve for $\mu$.Taking equation (3) and substituting for the value of $\sigma=19.38$ gives,

$\mu=40+0.5244005(19.38)=40+10.163=50.16(2dp)$

Therefore, the mean and standard deviation are 50.16 and 19.38 respectively.