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Check appedix of your probability or statistics textbook:
Given that Z follows standard normal distribution, then
1. P(Z<k) = 0.9177 => k=1.3898
2. P(Z>k) = 0.3192 => k=0.4699
3. P(0.4<Z<k) = 0.1997 => k=1.0586
X~N(73, 10^2)
a. P(X>98) = P(Z>(98-73)/10) = 0.0062
b. P(X<70) = P(Z<(70-73)/10) = 0.3821
c. P(82<X<91) = P((82-73)/10 < Z < (91-73)/10) =0.1218
X~N(10, 2^2):
a. P(X<13) = P(Z<(13-10)/2) = 0.9332
b. P(9<X≤12) = P((9-10)/2 < Z < (12-10)/2) = 0.5328
Given that Z follows standard normal distribution, then
1. P(Z<k) = 0.9177 => k=1.3898
2. P(Z>k) = 0.3192 => k=0.4699
3. P(0.4<Z<k) = 0.1997 => k=1.0586
X~N(73, 10^2)
a. P(X>98) = P(Z>(98-73)/10) = 0.0062
b. P(X<70) = P(Z<(70-73)/10) = 0.3821
c. P(82<X<91) = P((82-73)/10 < Z < (91-73)/10) =0.1218
X~N(10, 2^2):
a. P(X<13) = P(Z<(13-10)/2) = 0.9332
b. P(9<X≤12) = P((9-10)/2 < Z < (12-10)/2) = 0.5328
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