A population of values has a normal distribution with \( \mu=195.5 \) and \( \sigma=25.3 \). You intend to draw a random sample of size \( n=215 \). Find the probability that the sample mean is greater than 197.2.
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I apologize for the oversight. Let's solve it and provide the answer.
Step-by-step explanation :
1. **Standard Error Calculation :
\[ SE = \frac{\sigma}{\sqrt{n}} \]
\[ SE = \frac{25.3}{\sqrt{215}} \]
\[ SE \approx 1.7292 \]
2. **Z-Score Calculation:**
\[ z = \frac{X - \mu}{SE} \]
\[ z = \frac{197.2 - 195.5}{1.7292} \]
\[ z \approx 0.9836 \]
3. Finding the Probability :
To find the probability that the sample mean is greater than 197.2, we need to find the area to the right of the z-score in the standard normal distribution table.
For \( z \approx 0.9836 \), the area to the left is approximately \( 0.8374 \).
Since we want the area to the right (the probability the sample mean is greater than 197.2), we need to subtract that value from 1 :
\[ P(X > 197.2) = 1 - 0.8374 \]
\[ P(X > 197.2) = 0.1626 \]
Answer : The probability that the sample mean is greater than 197.2 is approximately \( 0.1626 \) or \( 16.26\% \).
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Step-by-step explanation:
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Answer:
C
Step-by-step explanation:
Let's say the cost of one sandwich is "s" and the cost of one drink is "d". From the first customer's order, we know that 3 sandwiches and 2 drinks cost $14.70. So we can write the equation: 3s + 2d = 14.70 From the second customer's order, we know that 2 sandwiches and 4 drinks cost $13.30. So we can write the equation: 2s + 4d = 13.30 Now, we can solve this system of equations to find the values of "s" and "d". Multiplying the first equation by 2 and the second equation by 3, we get: 6s + 4d = 29.40 6s + 12d = 39.90 Subtracting the first equation from the second equation, we get: 6s + 12d - (6s + 4d) = 39.90 - 29.40 Simplifying, we have: 8d = 10.50 Dividing both sides by 8, we find: d = 1.3125 Now we can substitute this value back into either of the original equations to find the value of "s". Let's use the first equation: 3s + 2(1.3125) = 14.70 Simplifying, we have: 3s + 2.625 = 14.70 Subtracting 2.625 from both sides, we find: 3s = 12.075 Dividing both sides by 3, we get: s = 4.025 So the cost of one sandwich is approximately $4.03 and the cost of one drink is approximately $1.31. Therefore, the correct answer is: c) Sandwich: $4.03, Drink: $1.31
Final answer:
Option (a), with the cost of a sandwich as $3.50 and a drink as $2.35, is the correct solution for this algebraic problem. This conclusion was reached by forming two equations from the information given and solving this system of equations.
Explanation:
This is an algebra problem where we set up two equations to solve for two variables. Let's denote the cost of a sandwich as S and the cost of a drink as D. The first equation derived from the first customer's purchase would be 3S + 2D = 14.70. The second equation from the second customer's purchase would be 2S + 4D = 13.30. To solve these equations, we could multiply the first equation by 2 and the second equation by 3 then subtract the second equation from the first. This will provide the cost of a Sandwich which can then be substituted back into either original equation to get the cost of a Drink. Once you solve this system, the answer appears as option (a): Sandwich $3.50 and Drink $2.35.
Learn more about System of equations here:
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Read more about correlation at:
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Hope I helped! Feel free to respond to my answer if anything is unclear.
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