document.write( "Question 927547: help Im am completely lost!
\n" ); document.write( "weights of a certain model of fully loaded gravel trucks follow normal distribution with mean u=6.4 tons and standard deviation o=0.3 tons. what is the probability that a fully loaded gravel truck of this model is
\n" ); document.write( "a) less than 6 tons?
\n" ); document.write( "b) more than 7 tons?
\n" ); document.write( "c) between 6 and 7 tons? \n" ); document.write( "
Algebra.Com's Answer #563109 by rothauserc(4718)\"\" \"About 
You can put this solution on YOUR website!
We are given a normal distribution with mean u=6.4 tons and standard deviation o=0.3 tons. In order to calculate probability (Pr), we must compute z-values and their associated Pr's.
\n" ); document.write( "note that z-value = (X value - mean value) / standard deviation, then consult the table of z-values to determine the associated Pr.
\n" ); document.write( "a) z-value = (6 - 6.4) / 0.3 = −1.333333333
\n" ); document.write( "Pr(X<6) = 0.0918
\n" ); document.write( "b) Pr(X>7) = 1 - Pr(X<7)
\n" ); document.write( "z-value = (7 - 6.4) / 0.3 = 2
\n" ); document.write( "Pr(X<7) = 0.9772
\n" ); document.write( "Pr(X>7) = 1 - 0.9772 = 0.0228
\n" ); document.write( "c) Pr( 6 < X < 7 ) = Pr(X<7) - Pr(X<6)
\n" ); document.write( "Pr( 6 < X <7 ) = 0.9772 - 0.0918 = 0.8854\r
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