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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 \( P(-2 < Z < 1) \approx 0.8185 \)
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