SOLUTION: Given a normal distribution with mean = 100 and standard deviation = 10, what is the probability that
a. X>70?
b. X<75?
c. X<90 OR X>125?
d. 99% of the values are between wh
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Question 1190823: Given a normal distribution with mean = 100 and standard deviation = 10, what is the probability that
a. X>70?
b. X<75?
c. X<90 OR X>125?
d. 99% of the values are between what two X-values (symmetrically distributed around the mean)?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
x is >70 is z>-3, when z=(x-mean)/sd
prob. z>-3=0.9987
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x<75 is z<-2.5 and that probability is from the table or calculator 2ndVARS2normalcdf(-6,-2.5)=0.0062. Any large negative number may be on the left side of the parentheses and any large positive number (6 is fine) may be on the right. Some use 1E99, but 6 is easier for me.
-
x<90 has probability of z<-1 and x>125 is z>2.5
=0.1587+0.0062
=0.1649
-
use the 0.5th and the 99.5th percentile.
z(0.005) use invnorm(0.005,0,1)=-2.576
z(0.995 is the positive number, since the function is symmetrical
because z=(x-mean)/sd
z*sd=x-mean=25.76
the answer is (74.24, 125.76)
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