To calculate the probability that a random variable \( x \) is between 3.6 and 6.1, given that the mean \( \mu = 5 \) and standard deviation \( \sigma = 0.5 \), we can standardize the variable using the Z-score formula:
\[ Z = \frac{X - \mu}{\sigma} \]
- Calculate the Z-scores for 3.6 and 6.1:
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For \( x = 3.6 \): \[ Z_{3.6} = \frac{3.6 - 5}{0.5} = \frac{-1.4}{0.5} = -2.8 \]
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For \( x = 6.1 \): \[ Z_{6.1} = \frac{6.1 - 5}{0.5} = \frac{1.1}{0.5} = 2.2 \]
- Find the probabilities corresponding to these Z-scores using the standard normal distribution table or a calculator:
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The probability for \( Z = -2.8 \): \[ P(Z < -2.8) \approx 0.0026 \] (This value can vary slightly depending on the table used, but it's around this ballpark.)
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The probability for \( Z = 2.2 \): \[ P(Z < 2.2) \approx 0.9861 \]
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Calculate the probability that \( x \) is between 3.6 and 6.1: \[ P(3.6 < X < 6.1) = P(Z < 2.2) - P(Z < -2.8) = 0.9861 - 0.0026 = 0.9835 \]
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Convert the probability to percentage: \[ 0.9835 \times 100 \approx 98.35% \]
So rounding gives us \( 98.4% \).
The final response is:
98.4%