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.

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.

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:

Where:

Here, you have:

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.

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:

describing distributions is one of the key skills you’ll need to earn a high score on the ap statistics exam. if you need proof of this, just flip through some past exam questions, which can be found at the collegeboard website. you’ll notice that the first free response question is almost always a question requiring you to look at a graph and describe it.

step-by-step explanation: just trying to get points sorry : (

i think the correct option is the second option.