Question

Suppose that the weight of seedless watermelons is normally distributed with mean 6.6 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.

What is the probability that a randomly selected watermelon will weigh more than 6 kg?

What is the probability that a randomly selected seedless watermelon will weigh between 6.2 and 7.1 kg?

The 90th percentile for the weight of seedless watermelons is

Answer 1

Solution:
Given in the question
Mean = 6.6
Standard deviation = 1.4
Solution(a)
We need to calculate P(X>6kg)
Z = (6-6.60)/1.4 = -0.43
So from Z table we found p-value
P(X>6kg) = 1- P(X<=6) = 1- 0.3335 = 0.6684
Solution(b)
P(6.2<X<7.1) = P(X<7.1) - P(X<6.2)
Z = (6.2-6.6)/1.4 = -0.2857
Z = (7.1-6.6)/1.4 = 0.3571
From Z table we found p-values
P(6.2<X<7.1) = P(X<7.1) - P(X<6.2) = 0.6406 - 0.3859 = 0.2547
Solution(c)
P-value = 0.9, Z-Score = 1.28
1.28 = (X-6.6)/1.4
1.792 = X-6.6
X = 6.6+1.792 = 8.39
So 90% for the weight of seedless watermelon is 8.39kg