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x is a normally distributed random variable with mean 49 and standard d…

Question

x is a normally distributed random variable with mean 49 and standard deviation 16. what is the probability that x is between 1 and 81? use the 0.68 - 0.95 - 0.997 rule and write your answer as a decimal. round to the nearest thousandth if necessary.

Explanation:

Step1: Calculate z-scores for 1 and 81

For $x=1$: $z_1 = \frac{1 - 49}{16} = -3$
For $x=81$: $z_2 = \frac{81 - 49}{16} = 2$

Step2: Apply empirical rule segments

  • Probability for $z=-3$ to $z=0$: $\frac{0.997}{2} = 0.4985$
  • Probability for $z=0$ to $z=2$: $\frac{0.95}{2} = 0.475$

Step3: Sum the two probabilities

$0.4985 + 0.475 = 0.9735$

Step4: Round to nearest thousandth

$0.9735 \approx 0.974$

Answer:

0.974