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Question 1182546: Suppose that the weight of seedless watermelons is normally distributed with mean 6.7 kg. and standard deviation 1.8 kg.
Let X be the weight of a randomly selected seedless watermelon.
Round all answers to 4 decimal places.
a. What is the distribution of X?
X -N(6.7 1.8)
b. What is the median seedless watermelon weighing 7.8 kg?
c. What is the Z-score for a seedless watermelon weighing 7.8 kg?
d. What is the possibility that a randomly selected watermelon will weigh more than 6.2 kg?
e. What is the probability that a randomly selected seedless watermelon will weigh between 6.3 and 7 kg?
f. The 70th percentile of the weight of seedless watermelons is _____ kg.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Use sigma^2 for the distribution, since it is the variance, so X~N(6.7, 3.24)
b. is not clear to me. The median weight is 6.7 kg, same as the mean for a normal distriution.
c. The median seedless watermelon weighing 7.8 kg from this distribution has a z-score of
(7.8-6.7)/1.8=+0.6111
d. z > (6.2-6.7)/1.8 which is >-0.2778, and that z-score means a probability of 0.6094
e. z > (-0.4/1.8)=-0.222 and z < (0.3/1.8)=0.166667
that probability is 0.1541
f. z is 0.5244
0.5244=(x-6.7)/1.8
0.9439=x-6.7
x=7.6439 kg
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