Answer to Question #222174 in Statistics and Probability for pesky

Question

500 candidates took the statistics exam in 2010 and their mean mark was found to be 58 with a standard deviation of 5.8.

i. Estimate the passmark if it is desired that only 300 candidates should succeed.

ii. Estimate the number if candidates who scored between 50 and 70 inclusive.

Answer 1

i. $\frac{300}{500}=0.6.$

$P(Z<z)=0.4.$

$z=-0.25.$

$\frac{x-58}{5.8}=-0.25.$

$x=58-0.25*5.8=56.55.$

Pass mark: 56.55.

ii. $P(50<X<70)=P(\frac{50-58}{5.8}<Z<(\frac{70-58}{5.8})=(-1.38<Z<2.07)=$

$=P(Z<2.07)-P(Z<-1.38)=0.8970.$

$N=500*0.8970\approx449.$