Question

Suppose that the weight of seedless watermelons is normally distributed with mean 6.1 kg. and standard deviation 1.2 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?

c. What is the Z-score for a seedless watermelon weighing 7 kg?

d. What is the probability that a randomly selected watermelon will weigh more than 5.1 kg?

e. What is the probability that a randomly selected seedless watermelon will weigh between 5.3 and 6 kg?

f. The 85th percentile for the weight of seedless watermelons is

Answer 1

a) N(6.1,1.2)
b) 6.1
c) 0.75
d) 0.7977 (or 79.77%)
e) 0.2143 (or 21.43%)
f) 7.3435 kg (using excel) or 7.348 kg (using normal tables)

Explanation:

a) normal distribution just need to be define we his mean and his standard deviation. You just need a sample higher than 30 or to be specified the population is normally distributed
X≈N(μ,σ)= N(6.1,1.2)

b) The median and mean are not necessarily the same, for a normal distribution, which is a symmetric distribution, those values are just the same


c) The equation for Z score is given by:

`z=(x-u)/(\sigma)`

Replacing values:

`Z=(7-6.1)/(1.2) =0.75`

d) first find the Zvalue for 5.1

`Z=(5.1-6.1)/(1.2) =-0.83`

Now, find the probability from Normal Distribution table
P(Z>0.83) = 0.7977

e)

first find the Zvalue fo5.3 and 6

`Z=(5.3-6.1)/(1.2) =-0.67`


`Z=(6-6.1)/(1.2) =-0.08`

Now find probabilities from Normal Distribution table . Notice you need to subtract P(Z>0.08)-P(Z>0.67) if you use a positive table.

P(-0.67<z<-0.08) = 0.2143


f) Find the Z score from a Normal Distribution table that give you an area of 0.8500 (or the closest value), im using excel for this one since the answer from tables have only TWO decimals and can be problematic if you need more decimal places.

this gives a Z of 1.036433389

now use the Zscore equation but solve for X

`z\sigma+u=x`

x = 1.036433389(1.2)+6.1 = 7.3437 (Using excel)

If i use a table, the closes is 1.04 (0.8508 which is not exactly 0.85)

x = 1.04(1.2)+6.1= 7.348 (using table)

so be carefully with the input if the website needs the decimals places or not.

Answer 2

a. X ~ N(6.1, 1.2^2)

b. The median seedless watermelon weight is 6.1 kg.

c. The Z-score for a seedless watermelon weighing 7 kg is 0.75.

d. The probability that a randomly selected watermelon will weigh more than 5.1 kg is 0.7967.

e. The probability that a randomly selected seedless watermelon will weigh between 5.3 and 6 kg is 0.2454.

f. The 85th percentile for the weight of seedless watermelons is 7.2437 kg.

Explanation:

a. X ~ N(6.1, 1.2^2)

b. The median of a normal distribution is equal to the mean, so the median seedless watermelon weight is 6.1 kg.

c. The Z-score for a seedless watermelon weighing 7 kg can be calculated as:

Z = (7 - 6.1) / 1.2 = 0.75

Therefore, the Z-score is 0.75.

d. To find the probability that a randomly selected watermelon will weigh more than 5.1 kg, we need to standardize the value using the formula:

Z = (X - μ) / σ

where X is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.

Z = (5.1 - 6.1) / 1.2 = -0.8333

Using a standard normal distribution table or a calculator, we can find the probability that Z is greater than -0.8333 to be 0.7967.

Therefore, the probability that a randomly selected watermelon will weigh more than 5.1 kg is 0.7967.

e. To find the probability that a randomly selected seedless watermelon will weigh between 5.3 and 6 kg, we need to standardize the values and find the area under the normal curve between the two Z-scores. The Z-scores for 5.3 kg and 6 kg are:

Z1 = (5.3 - 6.1) / 1.2 = -0.6667

Z2 = (6 - 6.1) / 1.2 = -0.0833

Using a standard normal distribution table or a calculator, we can find the probability that Z is between -0.6667 and -0.0833 to be 0.2454.

Therefore, the probability that a randomly selected seedless watermelon will weigh between 5.3 and 6 kg is 0.2454.

f. The 85th percentile for the weight of seedless watermelons can be found by finding the Z-score that corresponds to the 85th percentile of a standard normal distribution. Using a standard normal distribution table or a calculator, we can find the Z-score to be 1.0364.

To find the corresponding weight, we can use the formula:

Z = (X - μ) / σ

1.0364 = (X - 6.1) / 1.2

X = 7.2437

Therefore, the 85th percentile for the weight of seedless watermelons is 7.2437 kg.

Hope this helps! I'm sorry if it doesn't. If you need more help, ask me! :]