The mean score for freshmen on an aptitude test at a certain college is 540, with a standard deviation of 50. Assume the means to be measured to any degree of accuracy. What is the probability that two groups selected at random, consisting of 32 and 50 students, respectively, will differ in their mean scores by (a) more than 20 points

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

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )