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XX is a normally distributed random variable with mean 4545 and standard deviation 1010. What is the probability that XX is between 3535 and 5555? Use the 0.68−0.95−0.9970.68-0.95-0.997 rule and write your answer as a decimal. Round to the nearest thousandth if necessary.

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Q. XX is a normally distributed random variable with mean 4545 and standard deviation 1010. What is the probability that XX is between 3535 and 5555? Use the 0.68−0.95−0.9970.68-0.95-0.997 rule and write your answer as a decimal. Round to the nearest thousandth if necessary.
  1. Calculate Z-score for X=35X=35: We got a mean (μ\mu) of 4545 and a standard deviation (σ\sigma) of 1010. Let's find the z-score for X=35X=35.\newlineZ=X−μσ=35−4510=−1010=−1Z = \frac{X - \mu}{\sigma} = \frac{35 - 45}{10} = \frac{-10}{10} = -1.
  2. Calculate Z-score for X=55X=55: Now let's do the same for X=55X=55.Z=X−μσ=55−4510=1010=1Z = \frac{X - \mu}{\sigma} = \frac{55 - 45}{10} = \frac{10}{10} = 1.
  3. Find Probability for 35<x<5535<x<55:< b=""> So we're looking for P(35 < X < 55), which is the same as P(\mu - \sigma < X < \mu + \sigma). According to the 0.680.68-9595-0.9970.997 rule, the probability that XX is within one standard deviation (μ±σ)(\mu \pm \sigma) of the mean is about 0.680.68.
  4. Find Probability for 35<x<5535<x<55:< b=""> So we're looking for P(35 < X < 55), which is the same as P(\mu - \sigma < X < \mu + \sigma). According to the 0.680.68-9595-0.9970.997 rule, the probability that XX is within one standard deviation (μ±σ)(\mu \pm \sigma) of the mean is about 0.680.68. Therefore, P(35 < X < 55) is approximately 0.680.68.

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