Test grades on the last statistics exam had a mean = 78 and standard deviation = .14. suppose the teacher decides to curve by subtraction 31 from all scores then doubling the values. if y represents the new test scores, what is the mean and standard deviation of y?

The mean and standard deviation of Y are, 94 and 0.28 respectively.

Step-by-step explanation:

Let the random variable , 'Test grades on the last statistics exam' be X.

Then according to the question,

E(X) = 78 ------------(1)

and

= 0.14------------(2)

Now, according to the question,

Y = 2(X - 31)

⇒E(Y) = 2(E(X) - 31)

          =

          = 94  ----------------(4)

and

V(Y) = 4V(X)

⇒

⇒ = 0.28

So, the mean and standard deviation of Y are, 94 and 0.28 respectively.

an = 22 + (n-1) *6

110 seats

step-by-step explanation:

an=a1+(n-1)(d)

a1 = 22   this is the first item in the sequence

d =6   this is how the sequence goes up

a)   an = 22 + (n-1) *6

b) a12 = 22 + (12-1) *6

    a12 = 22+11*6

  a12 = 22 +88

  a12 = 110

2y+y=5

add 2y and y