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XX is a normally distributed random variable with mean 88 and standard deviation 55. What is the probability that XX is between 33 and 1313? 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 88 and standard deviation 55. What is the probability that XX is between 33 and 1313? 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. Identify Mean and Standard Deviation: Identify the mean (μ\mu) and standard deviation (σ\sigma) of the normal distribution.\newlineμ=8\mu = 8, σ=5\sigma = 5
  2. Calculate Z-score for X=3X = 3: Calculate the Z-score for X=3X = 3.Z=Xμσ=385=55=1Z = \frac{X - \mu}{\sigma} = \frac{3 - 8}{5} = \frac{-5}{5} = -1
  3. Calculate Z-score for X=13X = 13: Calculate the Z-score for X=13X = 13.Z=Xμσ=1385=55=1Z = \frac{X - \mu}{\sigma} = \frac{13 - 8}{5} = \frac{5}{5} = 1
  4. Use Probability Rule: Use the 0.680.68-0.950.95-0.9970.997 rule to find the probability that XX is between 33 and 1313. Since the ZZ-scores for X=3X = 3 and X=13X = 13 are 1-1 and 0.950.9500, respectively, this corresponds to the probability within one standard deviation of the mean, which is approximately 0.680.68.
  5. Write Final Answer: Write the final answer as a decimal, rounded to the nearest thousandth if necessary.\newlineThe probability that XX is between 33 and 1313 is approximately 0.680.68.

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