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XX is a normally distributed random variable with mean 1010 and standard deviation 33. What is the probability that XX is between 99 and 1111? Write your answer as a decimal rounded to the nearest thousandth.

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Q. XX is a normally distributed random variable with mean 1010 and standard deviation 33. What is the probability that XX is between 99 and 1111? Write your answer as a decimal rounded to the nearest thousandth.
  1. Calculate z-score for X=9X=9: Mean (μ\mu) is 1010, standard deviation (σ\sigma) is 33. Calculate the z-score for X=9X=9.\newlineZ=Xμσ=9103=13Z = \frac{X - \mu}{\sigma} = \frac{9 - 10}{3} = -\frac{1}{3}.
  2. Calculate z-score for X=11X=11: Calculate the z-score for X=11X=11.Z=Xμσ=11103=13Z = \frac{X - \mu}{\sigma} = \frac{11 - 10}{3} = \frac{1}{3}.
  3. Probability within 11 standard deviation: Using the 0.680.68-0.950.95-0.9970.997 rule, the probability that XX is within 11 standard deviation (μ±σ\mu \pm \sigma) is about 0.680.68.
  4. Probability between 99 and 1111: Since 99 and 1111 are within 11 standard deviation from the mean, the probability that XX is between 99 and 1111 is less than 0.680.68.
  5. Estimate exact probability: The exact probability for z-scores 13-\frac{1}{3} and 13\frac{1}{3} is not given by the 0.680.68-0.950.95-0.9970.997 rule, but we can estimate it to be roughly half of 0.680.68, because the interval from 99 to 1111 is about half the interval from μσ\mu - \sigma to μ+σ\mu + \sigma.

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