document.write( "Question 1182546: Suppose that the weight of seedless watermelons is normally distributed with mean 6.7 kg. and standard deviation 1.8 kg.
\n" ); document.write( "Let X be the weight of a randomly selected seedless watermelon.
\n" ); document.write( "Round all answers to 4 decimal places.\r
\n" ); document.write( "\n" ); document.write( "a. What is the distribution of X?
\n" ); document.write( " X -N(6.7 1.8)\r
\n" ); document.write( "\n" ); document.write( "b. What is the median seedless watermelon weighing 7.8 kg?\r
\n" ); document.write( "\n" ); document.write( "c. What is the Z-score for a seedless watermelon weighing 7.8 kg?\r
\n" ); document.write( "\n" ); document.write( "d. What is the possibility that a randomly selected watermelon will weigh more than 6.2 kg?\r
\n" ); document.write( "\n" ); document.write( "e. What is the probability that a randomly selected seedless watermelon will weigh between 6.3 and 7 kg?\r
\n" ); document.write( "\n" ); document.write( "f. The 70th percentile of the weight of seedless watermelons is _____ kg. \n" ); document.write( "

Algebra.Com's Answer #812643 by Boreal(15235) $\"\"$ $\"About$

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Use sigma^2 for the distribution, since it is the variance, so X~N(6.7, 3.24)
\n" ); document.write( "b. is not clear to me. The median weight is 6.7 kg, same as the mean for a normal distriution.
\n" ); document.write( "c. The median seedless watermelon weighing 7.8 kg from this distribution has a z-score of
\n" ); document.write( "(7.8-6.7)/1.8=+0.6111
\n" ); document.write( "d. z > (6.2-6.7)/1.8 which is >-0.2778, and that z-score means a probability of 0.6094
\n" ); document.write( "e. z > (-0.4/1.8)=-0.222 and z < (0.3/1.8)=0.166667
\n" ); document.write( "that probability is 0.1541
\n" ); document.write( "f. z is 0.5244
\n" ); document.write( "0.5244=(x-6.7)/1.8
\n" ); document.write( "0.9439=x-6.7
\n" ); document.write( "x=7.6439 kg\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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