Suppose that the weight of seedless watermelons is normally distributed with mean 6.6 kg. and standard deviation 1.1 kg. Let X be the weight of a randomly selected seedless watermelon. Round all answers to 4 decimal places where possible.

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d. What is the probability that a randomly selected watermelon will weigh more than 5.9 kg?
e. What is the probability that a randomly selected seedless watermelon will weigh between 6 and 6.8 kg?
f. The 75th percentile for the weight of seedless watermelons is
kg.
Transcribed Image Text:d. What is the probability that a randomly selected watermelon will weigh more than 5.9 kg? e. What is the probability that a randomly selected seedless watermelon will weigh between 6 and 6.8 kg? f. The 75th percentile for the weight of seedless watermelons is kg.
Suppose that the weight of seedless watermelons is normally distributed with mean 6.6 kg. and standard
deviation 1.1 kg. Let X be the weight of a randomly selected seedless watermelon. Round all answers to 4
decimal places where possible.
Transcribed Image Text:Suppose that the weight of seedless watermelons is normally distributed with mean 6.6 kg. and standard deviation 1.1 kg. Let X be the weight of a randomly selected seedless watermelon. Round all answers to 4 decimal places where possible.
Expert Solution
Step 1: Introducing the given information

Given information:

X denotes the weight of a randomly selected seedless watermelon.

Mean open parentheses mu close parentheses equals 6.6 kg

Standard deviation open parentheses sigma close parentheses equals 1.1 kg

Therefore,

X tilde N open parentheses mu equals 6.6 comma space sigma to the power of blank equals 1.1 to the power of blank close parentheses

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