2 Suppose that the weight of a paperback book is normally di
Solution
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 11.5
u = mean = 8
s = standard deviation = 1.5
Thus,
z = (x - u) / s = 2.333333333
Thus, using a table/technology, the right tailed area of this is
P(z > 2.333333333 ) = 0.009815329 [ANSWER]
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b)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 10.3
x2 = upper bound = 14
u = mean = 8
s = standard deviation = 1.5
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = 1.533333333
z2 = upper z score = (x2 - u) / s = 4
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.937403127
P(z < z2) = 0.999968329
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.062565202 [ANSWER]
