Question

Suppose that the weight of seedless watermelons is normally distributed with mean 6 kg. and standard deviation 1.6 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 6.6 kg?

d. What is the probability that a randomly selected watermelon will weigh more than 5.6 kg?

e. What is the probability that a randomly selected seedless watermelon will weigh between 5.1 and 5.8 kg?

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

Answer 1

a.
X ~ N(6, 1.6^2)

b)
median = 6

c)
z = (x - μ)/σ
z = (6.6 - 6)/1.6 = 0.38

d)
z = (x - μ)/σ
z = (5.6 - 6)/1.6 = -0.25

Therefore,
P(X >= 5.6) = P(z <= (5.6 - 6)/1.6)
= P(z >= -0.25)
= 1 - 0.4013 = 0.5987

e)
z = (x - μ)/σ
z1 = (5.1 - 6)/1.6 = -0.56
z2 = (5.8 - 6)/1.6 = -0.13

Therefore, we get
P(5.1 <= X <= 5.8) = P((5.8 - 6)/1.6) <= z <= (5.8 - 6)/1.6)
= P(-0.56 <= z <= -0.13) = P(z <= -0.13) - P(z <= -0.56)
= 0.4483 - 0.2877
= 0.1606

f)
z-value = 0.67
x = 6 + 0.67*1.6 = 7.0720 kg