Respuesta :
Answer:
P ( (X-Y) > |20| ) = 0.0772
Step-by-step explanation:
Solution:-
- Denote a random variable (X) with mean ux = 540 and standard deviation sx = 50 describes the sample of first group of nx = 32 students from the observed population.
- Denote a random variable (Y) with mean uy = 540 and standard deviation sy = 50 describes the sample of first group of ny = 50 students from the observed population.
- The sampling distribution is approximated normal ( X - Y ), where:
u(x-y) = ux - uy = 0
s(x-y) = √( sx^2/nx + sy^2/ny) = √( 50^2/32 + 50^2/50) = 11.319
- The distribution of difference in samples is given as:
(X-Y) ~ ( 0 , 11.319^2)
a)
- For the difference to be greater than 20.
- We will standardize our distribution (X-Y) to Z-statistics:
P ( (X-Y) > |20| ) = P ( (X-Y) < -20 ) + P ( (X-Y) > 20 )
= P ( Z < (-20 - 0) / 11.319) + P (Z > (20 - 0) / 11.319)
= P ( Z < -1.767) +1 - P (Z < 1.767)
= 0.0386 + 1 - 0.9614
= 0.0772
