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XX is a normally distributed random variable with mean 4444 and standard deviation 1313. What is the probability that XX is between 55 and 8383? Use the 0.680.950.9970.68-0.95-0.997 rule and write your answer as a decimal. Round to the nearest thousandth if necessary.

Full solution

Q. XX is a normally distributed random variable with mean 4444 and standard deviation 1313. What is the probability that XX is between 55 and 8383? Use the 0.680.950.9970.68-0.95-0.997 rule and write your answer as a decimal. Round to the nearest thousandth if necessary.
  1. Calculate z-score for X=5X=5: Mean (μ\mu) is 4444 and standard deviation (σ\sigma) is 1313. Calculate the z-score for X=5X=5.\newlineZ=Xμσ=54413=3913=3Z = \frac{X - \mu}{\sigma} = \frac{5 - 44}{13} = \frac{-39}{13} = -3.
  2. Calculate z-score for X=83X=83: Calculate the z-score for X=83X=83.Z=Xμσ=834413=3913=3Z = \frac{X - \mu}{\sigma} = \frac{83 - 44}{13} = \frac{39}{13} = 3.
  3. Probability within standard deviations: Using the 0.680.68-0.950.95-0.9970.997 rule, the probability that XX is within 11 standard deviation (σ\sigma) of the mean (μ\mu) is 0.680.68, within 2σ2\sigma is 0.950.95, and within 0.950.9500 is 0.9970.997.
  4. Check z-scores within 3σ3\sigma: Since the z-scores for X=5X=5 and X=83X=83 are 3-3 and 33, respectively, this falls within 33 standard deviations from the mean.
  5. Calculate probability between 55 and 8383: Therefore, the probability that XX is between 55 and 8383 is approximately 0.9970.997.

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