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(b) An electronics company manufactures resistors that have a mean resistance of 100 ohms and a standard deviation of 10 ohms. The distribution of resistance is normal. Find the probability that a random sample of n = 75 resistors will have an average of fewer than 95 ohms. Given that P(Z > 2.5)?
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(b) An electronics company manufactures resistors that have a mean res...
Calculating the Probability of Average Resistance
To find the probability that a random sample of 75 resistors will have an average resistance of fewer than 95 ohms, we need to use the Central Limit Theorem since the sample size is large (n = 75).
Step 1: Calculate the Standard Error
The standard error (SE) of the mean is calculated using the formula:
SE = σ / sqrt(n)
where σ is the standard deviation and n is the sample size.
In this case, σ = 10 ohms and n = 75 resistors. So, SE = 10 / sqrt(75) ≈ 1.15 ohms.
Step 2: Find the Z-score
Next, we calculate the Z-score using the formula:
Z = (X̄ - μ) / SE
where X̄ is the sample mean, μ is the population mean, and SE is the standard error.
Given that μ = 100 ohms and X̄ = 95 ohms, we get:
Z = (95 - 100) / 1.15 ≈ -4.35
Step 3: Find the Probability
Now, we need to find the probability that Z > 2.5. Since the Z-score we calculated (-4.35) is much lower than 2.5, the probability of having an average resistance of fewer than 95 ohms is very close to 0.
Therefore, the probability that a random sample of 75 resistors will have an average resistance of fewer than 95 ohms is extremely low, given that P(Z > 2.5) is so small.
To find the probability that a random sample of 75 resistors will have an average resistance of fewer than 95 ohms, we need to use the Central Limit Theorem since the sample size is large (n = 75).
Step 1: Calculate the Standard Error
The standard error (SE) of the mean is calculated using the formula:
SE = σ / sqrt(n)
where σ is the standard deviation and n is the sample size.
In this case, σ = 10 ohms and n = 75 resistors. So, SE = 10 / sqrt(75) ≈ 1.15 ohms.
Step 2: Find the Z-score
Next, we calculate the Z-score using the formula:
Z = (X̄ - μ) / SE
where X̄ is the sample mean, μ is the population mean, and SE is the standard error.
Given that μ = 100 ohms and X̄ = 95 ohms, we get:
Z = (95 - 100) / 1.15 ≈ -4.35
Step 3: Find the Probability
Now, we need to find the probability that Z > 2.5. Since the Z-score we calculated (-4.35) is much lower than 2.5, the probability of having an average resistance of fewer than 95 ohms is very close to 0.
Therefore, the probability that a random sample of 75 resistors will have an average resistance of fewer than 95 ohms is extremely low, given that P(Z > 2.5) is so small.
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(b) An electronics company manufactures resistors that have a mean resistance of 100 ohms and a standard deviation of 10 ohms. The distribution of resistance is normal. Find the probability that a random sample of n = 75 resistors will have an average of fewer than 95 ohms. Given that P(Z > 2.5)?
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(b) An electronics company manufactures resistors that have a mean resistance of 100 ohms and a standard deviation of 10 ohms. The distribution of resistance is normal. Find the probability that a random sample of n = 75 resistors will have an average of fewer than 95 ohms. Given that P(Z > 2.5)? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about (b) An electronics company manufactures resistors that have a mean resistance of 100 ohms and a standard deviation of 10 ohms. The distribution of resistance is normal. Find the probability that a random sample of n = 75 resistors will have an average of fewer than 95 ohms. Given that P(Z > 2.5)? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for (b) An electronics company manufactures resistors that have a mean resistance of 100 ohms and a standard deviation of 10 ohms. The distribution of resistance is normal. Find the probability that a random sample of n = 75 resistors will have an average of fewer than 95 ohms. Given that P(Z > 2.5)?.
(b) An electronics company manufactures resistors that have a mean resistance of 100 ohms and a standard deviation of 10 ohms. The distribution of resistance is normal. Find the probability that a random sample of n = 75 resistors will have an average of fewer than 95 ohms. Given that P(Z > 2.5)? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about (b) An electronics company manufactures resistors that have a mean resistance of 100 ohms and a standard deviation of 10 ohms. The distribution of resistance is normal. Find the probability that a random sample of n = 75 resistors will have an average of fewer than 95 ohms. Given that P(Z > 2.5)? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for (b) An electronics company manufactures resistors that have a mean resistance of 100 ohms and a standard deviation of 10 ohms. The distribution of resistance is normal. Find the probability that a random sample of n = 75 resistors will have an average of fewer than 95 ohms. Given that P(Z > 2.5)?.
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