Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Our Q&A platform provides quick and trustworthy answers to your questions from experienced professionals in different areas of expertise. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly 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 was helpful. Please come back whenever you need more information or answers to your queries. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
