Suppose that the weight of seedless watermelons is normally distributed with mean 6.2 kg. and standard deviation 1.5 kg. Let X be the weight of a randomly selected seedless watermelon. Round all answers to two decimal places.

A. X ~ N(___ , )

B. What is the median seedless watermelon weight? kg.

C. What is the Z-score for a seedless watermelon weighing 8 kg?

D. What is the probability that a randomly selected watermelon will weigh more than 7 kg?

E. What is the probability that a randomly selected seedless watermelon will weigh between 4 and 5 kg?

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

A.

X ~ N(6.2kg, 2.25kg²)

B. What is the median seedless watermelon weight?____ kg.

6.2 kg

C. What is the Z-score for a seedless watermelon weighing 8 kg?

1.2

D. What is the probability that a randomly selected watermelon will weigh more than 7 kg?

0.2981

E. What is the probability that a randomly selected seedless watermelon will weigh between 4 and 5 kg?

0.1411

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

7.5 kg

Explanation:

A. X ~ N(___ , ____ )

The distribution of a random variable in a sample extracted from a population that follows a normal distribution is represented by the notation:

X ~ N(μ, σ²)

Where:

X is the random variableN stands for normal distribution functionμ is the median of the populationσ² is the variance of the population

Here, you have:

μ = 6.2 kgσ² = (1.5kg)² = 2.25 kg²X ~ N(6.2kg, 2.25kg²)

B. What is the median seedless watermelon weight?____ kg.

The median of a random variable that follows a normal distribution is equal to the mean, thus it is 6.2 kg.

C. What is the Z-score for a seedless watermelon weighing 8 kg?

The z-score is the standardized value of the random variable. It measures how far away is the variable from the mean.

It is calcuated with the formula:

       

Thus for X=8:

     

D. What is the probability that a randomly selected watermelon will weigh more than 7 kg?

You want P(X>7)

You must use the tables for the standardized normal distribution.

Find the corresponding Z-score for X = 7

       

You must use a table for the standardized normal distribution which gives the cumulative distribution or area under the curve of the standard normal distribution and find P(Z>0.53).

That is the area to the right of Z=0.53. The table shows 0.2981.

Thus, the probability that a randomly selected seedless watermelon will weigh more than 7 kg is 0.2981

E. What is the probability that a randomly selected seedless watermelon will weigh between 4 and 5 kg?

For this case you must find the Z-scores for X=4 and X=5 and then find the area under the curve of the standardized normal distribution between those two Z-scores.

Z(X=4) = (4 - 6.2)/1.5 ≈ -1.47Z(X=5) = (5 - 6.2)/1.5 = -0.8

In the table the area to the right of Z  = - 1.47 is 1 - 0.0708 = 0.9292

And the area to the right of Z = - 0.8 is 1 - 0.2119 = 0.7881

Thus, the area in between is the difference 0.9292 - 0.7881 = 0.1411.

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

The 80th percentile is the weigh of the top 20% seedless watermelons: this is 80% of the weighs are below that weight.

You must find the Z-score for which the area under the curve is less than 0.80.

The area less than 0.80 is 1 less the area that is less than 0.20.

From the table, the Zscore that defines the area less than 0.20 is 0.845 (interpolating).

Thus, the 80th percentile is the X value that makes the Z-score greater than or equal to 0.845:

Z ≥ 0.845(X - 6.2)/1.5 ≥ 0.845X ≥ 0.845 × 1.5 + 6.2X ≥ 7.4675X ≥ 7.5 kg ← answer

50.27 cm²

Step-by-step explanation:

Subtract the area of the smaller circle from the area of the larger circle

The area of a circle = πr² ( r is the radius )

area = π × 5² - π × 3² = 25π - 9π = 16π ≈ 50.27 cm² ( nearest hundredth )