This is not true, although both distributions contain x^(k-1) or x^(k/2-1) term before the exponent term. X-square exponent is e^(-x/2) whilst Weibull is e^(-x/lambda)^k. And for k=1 and lambda=2 Weibull distribution loses its (x/lambda)^(k-1) term
I thank the teacher very much on his effort but the English accent is not clear at all, please provide a subtitle for his talking it will be very helpful
