Answer: 0.9974
Step-by-step explanation:
Let X be the random variable that follows the normal distribution \.
If mean ([tex]\mu[/tex]) = 10 and standard deviation ([tex]\sigma[/tex]) =1
then, the probability of selecting a number between 7 and 13 = [tex]P(7<X<13)[/tex]
[tex]=P(\dfrac{7-10}{1}<\dfrac{X-\mu}{\sigma}<\dfrac{13-10}{1})[/tex]
[tex]=P(-3<z<3)[/tex] [Since [tex]z=\dfrac{X-\mu}{\sigma}[/tex]]
[tex]=P(z<3)-P(z<-3) \ \ \ [P(z_1<Z<z_2)=P(Z<z_2)-P(Z<z_1)]\\\\= P(z<3)-(1-P(z<3))\ \ \ [P(Z<-z)=1-P(Z<z)]\\\\=2P(z<3)-1\\\\=2(0.9987)-1\ \ [\text{By p-value table}]\\\\= 0.9974[/tex]
Hence, the probability of selecting a number between 7 and 13 is 0.9974.
