X is a normally distributed random variable with a mean of 60 and standard
deviation of 8. Find the probabilities indicated by using the table.
(a) P(X < 52)
(b) P(48 < X < 64)
(c) P(X > 57)
(a)P(X<52)=P(z<52−608)=P(z<−1)=0.5−P(0<z<1)=0.5−0.3413=0.1587(b)P(48<X<64)=P(48−608<z<64−608)=P(−1.5<z<0.5)=P(0<z<0.5)+P(0<z<1.5)=0.1915+0.4332=0.6247(c)P(X>57)=P(z>57−608)=P(z>−0.38)=0.5+P(0<z<0.38)=0.5+0.1480=0.6480(a) P(X < 52)=P(z < \frac{52-60}{8})\\=P(z < -1)=0.5-P(0<z < 1)\\=0.5-0.3413=0.1587 \\(b) P(48 < X < 64)=P(\frac{48-60}{8}<z < \frac{64-60}{8})\\=P(-1.5<z < 0.5)\\=P(0<z < 0.5)+P(0<z < 1.5)\\=0.1915+0.4332=0.6247 \\(c) P(X > 57)=P(z > \frac{57-60}{8})\\=P(z >-0.38)=0.5+P(0<z <0.38)\\=0.5+0.1480=0.6480(a)P(X<52)=P(z<852−60)=P(z<−1)=0.5−P(0<z<1)=0.5−0.3413=0.1587(b)P(48<X<64)=P(848−60<z<864−60)=P(−1.5<z<0.5)=P(0<z<0.5)+P(0<z<1.5)=0.1915+0.4332=0.6247(c)P(X>57)=P(z>857−60)=P(z>−0.38)=0.5+P(0<z<0.38)=0.5+0.1480=0.6480
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment