4-56. Assume that X is normally distributed with a mean of 5 and a standard deviation of 4. Determine the value for x that solves each of the following: (0) (a) P(X>x)=0.5 (c) P(xx)=0.95 (d) P(3

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Q.4-56
10)=0.99
4-55. Assume that X is normally distributed with a mean of 5
and a standard deviation of 4. Determine the following:
(a) P(X<11)
(d) P(-2<X<9)
e
(b) P(X>2)
(e) P(2<X<8)
(c) P(3<X<7)
4-56. Assume that X is normally distributed with a mean of
5 and a standard deviation of 4. Determine the value for x that
solves each of the following:
(a) P(X>x)=0.5
(c) P(x<X<7) = 0.2
(e) P(-x<X – 5<x)= 0.99
eplods
(b) P(X>x)=0.95
(d) P(3<x<x)= 0.95
A (1)
%3D
4-57. The compressive strength of samples of cement can be
modeled by a normal distribution with a mean of 6000 kilo-
grams per square centimeter and a standard deviation of 100
kilograms per square centimeter.
Transcribed Image Text:10)=0.99 4-55. Assume that X is normally distributed with a mean of 5 and a standard deviation of 4. Determine the following: (a) P(X<11) (d) P(-2<X<9) e (b) P(X>2) (e) P(2<X<8) (c) P(3<X<7) 4-56. Assume that X is normally distributed with a mean of 5 and a standard deviation of 4. Determine the value for x that solves each of the following: (a) P(X>x)=0.5 (c) P(x<X<7) = 0.2 (e) P(-x<X – 5<x)= 0.99 eplods (b) P(X>x)=0.95 (d) P(3<x<x)= 0.95 A (1) %3D 4-57. The compressive strength of samples of cement can be modeled by a normal distribution with a mean of 6000 kilo- grams per square centimeter and a standard deviation of 100 kilograms per square centimeter.
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