Find the following probability for the standard normal random variable z: A) P(-1 ≤ z ≤ 1)
B) P(-2 ≤ z ≤ 2)
C) P(-2.61 ≤ z ≤ 0.57)
D) P(-0.95< z < 1.11)

i)      P(-1 ≤ z ≤ 1) = 0.6826

ii)     P(-2 ≤ z ≤ 2) = 0.9544

iii)    P(-2.61 ≤ z ≤ 0.57) = 0.7112

iv)      P(-0.95< z < 1.11) = 0.6954

Step-by-step explanation:

i)

P(-1 ≤ z ≤ 1) = P(  z ≤ 1 ) - P(z≤-1)

                  =  0.5 + A(1) - (0.5 - A(-1))

                  =    0.5 + A(1) - 0.5 + A(-1)

                  = A(1) + A(1)     (∵ A(-1) = A(1)

                  = 2 × A(1)

                  = 2×0.3413     ( From Normal table)

P(-1 ≤ z ≤ 1) = 0.6826

ii)

P(-2 ≤ z ≤ 2) = P(  z ≤ 2 ) - P(z≤-2)

                  =  0.5 + A(2) - (0.5 - A(-2))

                  =    0.5 + A(2) - 0.5 + A(-2)

                  = A(2) + A(2)     (∵ A(-2) = A(2)

                  = 2 × A(2)

                 = 2 × 0.4772

                  = 0.9544

P(-2 ≤ z ≤ 2) = 0.9544

iii)

        P(-2.61 ≤ z ≤ 0.57) = P( z ≤ 0.57)-P(z ≤-2.61)

                                       = 0.5 + A(0.57) - (0.5 - A(-2.61))

                                        = A( 0.57) + A( 2.61)       ( ∵ A(-2.61) = A(2.61)

                                        = 0.2157 +0.4955

                                        = 0.7112

       P(-2.61 ≤ z ≤ 0.57) = 0.7112

iv)

      P(-0.95< z < 1.11) = P( z ≤ 1.11)-P(z ≤-0.95)

                                   = 0.5 + A(1.11) - (0.5 - A(-0.95))

                                   = A( 1.11) + A( 0.95)   (∵ A(-0.95) = A(0.95)

                                  = 0.3665 + 0.3289

                                  = 0.6954

       P(-0.95< z < 1.11) = 0.6954

$6.42

step-by-step explanation:

answer: 7. if the functions had a negative leading coefficient and is of even degree, which statement about the graph is true?

step-by-step explanation: