Question

Assume Z is a random variable with a standard normal distribution and c is a positive number. If P(Z > c) = 0.25, then PC – c< < c) = 0.5. O True OFalse

Answer 1

# 2 is a random variable with a standard normal dist and c is a position Number If poz>c) = 0.95 then. Pc-c<z<c) = P(Z <c) - P(Z <-C) = 42CC) - ºz(-C3 cere also know. z is symmetric so Pz Cac) + Paz (c) = 1 Plz G-C9 = 1 - 92CC) Hend. PC-czz<c) = 242CC) -1 = 2 (PCZ<c9}-/ *4 = 9{£- P(Z>()) - 1 = 2{1-0.25} -| = 9-0.50-1 = 1-0.50 20.50 So PC-cez<c) = 0.50 in Trul a # Exactly so% of the area under the normal cusne lied to the left of the meon - True Because Normal carnets symmetric. som 50% and meam = median.. yemen # PCX<5.2) = 0.5 then PCX> 5.2) : l-p(x<5.22 = -0-$ = 0.5 Hence, P(x>5.2) 20.5 is true.