Answer to Question #332046 in Statistics and Probability for Rival

Question

The grade of students are normally distributed with mean of 75 and the standard deviation of 5.

a) How many percent of student took the grade between 70 and 80?

b) How many percent of student took the grade between 65 and 85?

c) How many percent of student took the grade between 60 and 90?

d) How many percent of student took the grade less than 70?

e) How many percent of student took the grade between 60 and 80?

f) How many percent of student took the grade greater than 90?

g) How many percent of student took the grade between 65 and 75?

h) How many percent of student took the grade between 70 and 85?

Answer 1

a. $P(70<X<80)=P(\frac{70-75}{5}<Z<\frac{80-75}{5})=P(-1<Z<1)=0.8413-0.1587=0.6826$

b.

$P(65<X<85)=P(\frac{65-75}{5}<Z<\frac{85-75}{5})=P(-2<Z<2)=0.9772-0.0228=0.9544$

c.

$P(60<X<90)=P(\frac{60-75}{5}<Z<\frac{90-75}{5})=P(-3<Z<3)=0.9987-0.0013=0.9974$

d.$P(X<70)=P(Z<\frac{70-75}{5})=P(Z<-1)=0.1587$

e.

$P(60<X<80)=P(\frac{60-75}{5}<Z<\frac{80-75}{5})=P(-3<Z<1)=0.8413-0.0228=0.8185$

f.$P(X>90)=1-P(X<90)=1-P(Z<\frac{90-75}{5})=1-P(Z<3)=1-0.9772=0.0228$

g.$P(65<X<75)=P(\frac{65-75}{5}<Z<\frac{75-75}{5})=P(-2<Z<0)=0.5-0.1587=0.3413$

h. $P(70<X<85)=P(\frac{70-75}{5}<Z<\frac{85-75}{5})=P(-1<Z<2)=0.9772-0.1587=0.8185$