Question

Suppose that the weight of seedless watermelons is normally distributed with mean 6.2 kg. and standard deviation 1.4 kg. Let X be the weight of a randomly selected seedless watermelon. Round all answers to 4 decimal places where possible.

a. What is the distribution of X? X ~ N(___,___)

b. What is the median seedless watermelon weight? ______kg.

c. What is the Z-score for a seedless watermelon weighing 6.8 kg?

d. What is the probability that a randomly selected watermelon will weigh more than 7.2 kg?

e. What is the probability that a randomly selected seedless watermelon will weigh between 6.1 and 7 kg?

f. The 75th percentile for the weight of seedless watermelons is ______kg.

Answer 1

a)X~N(6.2 , 1.4)

b) median seedless watermelon weight =6.2

c) Z-score for a seedless watermelon weighing 6.8 kg =(6.8-6.2)/1.4 =0.4286

d)

probability =

P(X>7.2)

=

P(Z>0.71)=

1-P(Z<0.71)=

1-0.7625=

0.2375

(try 0.2389 if this comes wrong)_

e)

probability =

P(6.1<X<7)

=

P(-0.07<Z<0.57)=

0.7161-0.4715=

0.2446

( please try 0.2436 if this comes wrong)

f)

for 75th percentile critical value of z=		0.67
therefore corresponding value=mean+z*std deviation=			7.14