Assume you build a Candy castle for which you buy a bag of gummy bears, a block of chocolate and a bag of nuts. The weight of the three ingredients is a random variable (X1, X2 and X3) with mean weights μ1 = 200 gram, μ2 = 400 gram and μ3 = 150 gram. The bags and blocks are not measured top perfection so the standard deviation of the three is σ1 = 10, σ2 = 8 and σ3 = 15. The weight of the three ingredients is normally distributed!
a) Define the distribution of the weight of the candy castle!
b) What is the chance that the castle will weigh more than 700 gram?
c) What is the chance the castle will weigh between 730 and 755 gram?
d) With 75% chance the castle will weigh between ___ and ___. (symmetric range, as always)
e) The castle weighs more than how much with 95% chance?
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a) The distribution of the weight of the candy castle can be defined as the sum of the weights of the three ingredients: X = X1 + X2 + X3.
b) To find the chance that the castle will weigh more than 700 grams, we need to calculate the probability P(X > 700). Since the weights of the three ingredients are normally distributed, the sum of two or more normally distributed variables is also normally distributed. So, X will also be normally distributed.
To calculate this probability, we need to standardize the variable X and then use the standard normal distribution.
Standardizing X:
Z = (X - μ) / σ
Where μ is the mean of X (μ1 + μ2 + μ3) and σ is the standard deviation of X (sqrt(σ1^2 + σ2^2 + σ3^2)).
Substituting the given values:
Z = (700 - (200 + 400 + 150)) / sqrt(10^2 + 8^2 + 15^2)
Now, using a standard normal distribution table or a calculator, we can find the probability P(Z > Z-score) where Z-score is the standardized value obtained above.
c) To find the chance that the castle will weigh between 730 and 755 grams, we need to calculate the probability P(730 < X < 755).
Similar to part b, we need to standardize the variable X using the mean and standard deviation calculated in part a. Then we can find the probability P(Z-score1 < Z < Z-score2), where Z-score1 and Z-score2 are the standardized values for 730 and 755, respectively.
d) With a 75% chance, the castle will weigh between ___ and ___.
To find the symmetric range with a given percentage, we can use the concept of the standard normal distribution.
The 75th percentile corresponds to a Z-score of 0.674. We can find the corresponding weights using the formula:
Weight1 = μ - (Z-score * σ)
Weight2 = μ + (Z-score * σ)
Substituting the given values, we can find the weights for the symmetric range with a 75% chance.
e) To find the weight of the castle with a 95% chance, we need to find the Z-score corresponding to the 95th percentile of the standard normal distribution.
Using a standard normal distribution table or a calculator, we can find the Z-score, and then use it to calculate the weight of the castle using the formula:
Weight = μ + (Z-score * σ)
To answer these questions, we need to understand a few concepts related to probability distributions and the properties of normal distributions.
a) The distribution of the weight of the candy castle is a sum of the individual weights of the ingredients. According to the properties of normal distributions, the sum of independent normal random variables is also a normal random variable. Therefore, the weight of the candy castle follows a normal distribution.
b) To find the chance that the castle will weigh more than 700 grams, we need to calculate the probability of the candy castle's weight being greater than 700 grams.
The first step is to standardize the values using z-scores. The formula for the z-score is:
z = (x - μ) / σ
where x is the value of interest, μ is the mean, and σ is the standard deviation.
For the weight of the castle, the mean would be the sum of the means of the ingredients:
μ_candy = μ1 + μ2 + μ3 = 200 + 400 + 150 = 750 grams
The standard deviation of the sum of independent random variables is calculated by adding the variances:
σ_candy = sqrt(σ1^2 + σ2^2 + σ3^2) = sqrt(10^2 + 8^2 + 15^2)
Once we have the standardized values, we can find the cumulative probability using a standard normal distribution table or a calculator.
P(weight > 700 grams) = P(z > (700 - μ_candy) / σ_candy)
c) To find the chance that the castle will weigh between 730 and 755 grams, we need to calculate the cumulative probability for both values and subtract them.
P(730 ≤ weight ≤ 755) = P(z ≤ (755 - μ_candy) / σ_candy) - P(z ≤ (730 - μ_candy) / σ_candy)
d) With 75% chance, the castle will weigh between ___ and ___ (symmetric range).
To find the symmetric range, we need to find the corresponding z-scores for the percentiles. We can find these z-scores from the standard normal distribution table or a calculator.
Let's find the z-score for the lower bound of the 75th percentile (25th percentile).
z_lower = z-score for percentile = 0.25
Using the z-score formula mentioned earlier, we can solve for the corresponding value of the candy castle's weight (x):
x_lower = z_lower * σ_candy + μ_candy
Similarly, for the upper bound of the 75th percentile (75th percentile), we need to find the corresponding z-score:
z_upper = z-score for percentile = 0.75
x_upper = z_upper * σ_candy + μ_candy
e) To find the weight of the castle that exceeds 95% of the other castles, we need to find the corresponding z-score for the 95th percentile (5th percentile).
z = z-score for percentile = 0.05
Once we have the z-score, we can find the corresponding weight using the formula:
x = z * σ_candy + μ_candy