hw6sol - left of z is 0.90 This z is 1.28 Then we solve the...

Solutions to Homework 6
Problem 6.11
a. P ( z ≤ - 1 . 0) = 0 . 1587
b. P ( z ≥ - 1) = 1 - 0 . 1587 = 0 . 8413
c. P ( z ≥ - 1 . 5) = 1 - 0 . 0668 = 0 . 9332
d. P ( - 2 . 5 ≤ z ) = 1 - 0 . 0062 = 0 . 9938
e. P ( - 3 < z ≤ 0) = 0 . 5000 - . 0013 = 0 . 4987
Problem 6.13
a. P ( - 1 . 98 ≤ z ≤ 0 . 49) = 0 . 6879 - 0 . 0239 = 0 . 6640
b. P (0 . 52 ≤ z ≤ 1 . 22) = 0 . 8888 - 0 . 6985 = 0 . 1903
c. P ( - 1 . 75 ≤ z ≤ - 1 . 04) = 0 . 1492 - 0 . 0401 = 0 . 1091
Problem 6.15
a. z = - 0 . 80
b. If the area between z and - z is 0.9030, then the area less than - z and greater than z
is 1 - 0 . 9030 = 0 . 0970. So the area just less than - z is
0 . 0970
2
= 0 . 0485. So - z = - 1 . 66
and z = 1 . 66.
c. If the area between z and - z is 0.2052, then the area less than - z and greater than z
is 1 - 0 . 2052 = 0 . 7948. So the area just less than - z is
0 . 7948
2
= 0 . 3974. So - z = - 0 . 26
and z = 0 . 26.
d. z = 2 . 56
e. If the area to the right of z is 0.6915, then the area to the left of z is 1 - 0 . 6915 = 0 . 3085.
So z = - 0 . 50.
Problem 6.19
Here we have a normal distribution with μ = 3 . 5 inches and σ = 0 . 8 inches.
a. P ( x > 5) = P ( z > 5 - 3 . 5
0 . 8
) = P ( z > 1 . 8) = 1 - 0 . 9699 = 0 . 0301. So 3.01%.
b. P ( x < 3) = P ( z <
3 - 3 . 5
0 . 8
) = P ( z < - 0 . 63) = 0 . 2643. So 26.43%.
1

c. First, we need to find z such that the area to the right of