I'll be using the empirical (68-95-99.7) rule extensively. It also helps to know that the normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is symmetric about its mean.
a. [tex]P(22<X<25)=P(\mu<X<\mu+\sigma)=\dfrac{P(|X-\mu|<\sigma)}2\approx34\%[/tex]
b. [tex]P(16<X<28)=P(|X-\mu|<2\sigma)\approx95\%[/tex]
c. [tex]P(X<13)=P(X<\mu-3\sigma)=\dfrac{1-P(|X-\mu|<3\sigma)}2\approx1.5\%[/tex]
d. [tex]P(X>25)=P(X>\mu+\sigma)=\dfrac{1-P(|X-\mu|<\sigma)}2\approx16\%[/tex]
e. [tex]P(X>22)=50\%[/tex], so about 50 juniors scored above the average.
f. [tex]P(19<X<25)=P(|X-\mu|<\sigma)\approx68\%[/tex], so about 68 juniors scored between 19 and 25.
