Question

Suppose that the weight of seedless watermelons is normally distributed with mean 6.1 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 6.8 kg?

d. What is the probability that a randomly selected watermelon will weigh more than 5.8 kg?

e. What is the probability that a randomly selected seedless watermelon will weigh between 6 and 6.7 kg?

f. The 80th percentile for the weight of seedless watermelons is kg.

Answer 1

Here µ=6.1 and    σ=1.6

a)

Distribution of X is X~N ( µ , σ )

So X~ N ( 6.1 , 1.6 )

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b)

For Normal distribution Mean = Median =µ

So Median = 6.1 kg

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c)

The formula for Z score is

                    

So Z score is 0.4375

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d)

Probability that a randomly selected watermelon will weigh more than 5.8 kg

We will use Ti83/84

Press 2ND...VARS...normalcdf(

Enter values as

Result is

Probability that a randomly selected watermelon will weigh more than 5.8 kg is 0.5744

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e)

Probability that a randomly selected seedless watermelon will weigh between 6 and 6.7 kg

We will use Ti83/84

Press 2ND...VARS...normalcdf(

Enter values as

Result is

Probability that a randomly selected seedless watermelon will weigh between 6 and 6.7 kg is 0.1711

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f)

The 80th percentile for the weight of seedless watermelons

We will use Ti83/84

Press 2ND...VARS...invNorm(

Enter values as

Result is

The 80th percentile for the weight of seedless watermelons is 7.4466 kg.