Answer to Question #182193 in Statistics and Probability for Jayden

Question

Dhaka University took a test and the final grades have a mean of 100 and a standard deviation of 11. If we can approximate the distribution of these grades by a normal distribution, what percent of the students

a) scored higher than 120?

b) should pass the test (grades≥90)?

c) should fail the test (grades<90)?

Answer 1

Denote

$\overline{x}=100$

$s=11$

a) $P(X>120)=P(120<X<+\infty)=$

$F(+\infty)-F\left(\dfrac{120-\overline{x}}{s}\right)=$

$1-F\left(\dfrac{120-100}{11}\right)=$ $1-F\left(\dfrac{20}{11}\right)\approx$ $1-F\left(1.8182\right)\approx1-0.9655\approx0.0345$

Percent of the students scored higher than 120 is $0.0345\times100=3.45\%$

b) $P(X\geq90)=P(90\leq X<+\infty)=$

$F(+\infty)-F\left(\dfrac{90-\overline{x}}{s}\right)=$

$1-F\left(\dfrac{90-100}{11}\right)=$ $1-F\left(-\dfrac{10}{11}\right)\approx$

$1-F\left(-0.9091\right)\approx1-0.1817\approx0.8183$

Percent of the students should pass the test is $0.8183\times100=81.83\%$

c) $P(X<90)=P(-\infty<X<90)=$

$F\left(\dfrac{90-\overline{x}}{s}\right)-F(-\infty)=$

$F\left(\dfrac{90-100}{11}\right)-0=F\left(-\dfrac{10}{11}\right)\approx$ $F(-0.9091)\approx0.1817$

Percent of the students should fail the test is $0.1817\times100=18.17\%$

Note that b) and c) must satisfy $81.83\%+18.17\%=100\%$