Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
Answer:
Step-by-step explanation:
To address these questions, we'll use the properties of the sum of normally distributed variables. Given that the mean of the sugar amount is \( \mu = 39.01 \) and the standard deviation is \( \sigma = 0.5 \), each can's sugar content can be considered normally distributed.
For a sample of 100 cans:
- The mean of the sum \( S_{100} \) of these 100 cans is \( \mu_{S_{100}} = 100 \times 39.01 = 3901 \).
- The standard deviation of the sum \( S_{100} \) is \( \sigma_{S_{100}} = \sqrt{100} \times 0.5 = 5 \).
Now, let's calculate the probabilities:
**a) Probability that the sum of the 100 values is greater than 3,910:**
Convert this to a standard normal variable \( Z \):
\[ Z = \frac{3910 - 3901}{5} = \frac{9}{5} = 1.8 \]
Now, find the probability \( P(Z > 1.8) \) using the standard normal distribution table or calculator:
\[ P(Z > 1.8) \approx 0.0359 \]
**b) Probability that the sum of the 100 values is less than 3,900:**
Convert this to a standard normal variable \( Z \):
\[ Z = \frac{3900 - 3901}{5} = -0.2 \]
Now, find the probability \( P(Z < -0.2) \):
\[ P(Z < -0.2) \approx 0.4207 \]
**c) Probability that the sum of the 100 values falls between the numbers found in parts (a) and (b):**
This is \( P(3900 < S_{100} < 3910) \). Using the standard normal probabilities from parts (a) and (b):
\[ P(3900 < S_{100} < 3910) = 1 - [P(Z \leq -0.2) + P(Z \geq 1.8)] \]
\[ P(3900 < S_{100} < 3910) = 1 - [0.4207 + 0.0359] \]
\[ P(3900 < S_{100} < 3910) \approx 1 - 0.4566 \]
\[ P(3900 < S_{100} < 3910) \approx 0.5434 \]
**d) Probability that the sum of the 100 values falls between the z-scores of -2 and 1:**
Convert -2 and 1 to standard normal variables:
For \( Z = -2 \):
\[ P(Z < -2) \approx 0.0228 \]
For \( Z = 1 \):
\[ P(Z < 1) \approx 0.8413 \]
Now, find \( P(-2 < Z < 1) \):
\[ P(-2 < Z < 1) = P(Z < 1) - P(Z < -2) \]
\[ P(-2 < Z < 1) = 0.8413 - 0.0228 \]
\[ P(-2 < Z < 1) = 0.8185 \]
Therefore, the probabilities for the scenarios described are:
- **a)** Probability \( P(S_{100} > 3910) \approx 0.0359 \)
- **b)** Probability \( P(S_{100} < 3900) \approx 0.4207 \)
- **c)** Probability \( P(3900 < S_{100} < 3910) \approx 0.5434 \)
- **d)** Probability \( P(-2 < Z < 1) \approx 0.8185 \)
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
