\( X \) is a normally distributed random variable with mean 55 and standard deviation 18 . What is the probability that \( X \) is between 73 and 91 ? Use the \( 0.68-0.95-0.997 \) rule and write your answer as a decimal. Round to the nearest thousandth if necessary.
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To find the probability that \( X \) is between 73 and 91 using the \( 0.68-0.95-0.997 \) rule, we first standardize the values of \( X \) using the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] For \( X = 73 \): \[ Z(73) = \frac{73 - 55}{18} = 1.000 \] For \( X = 91 \): \[ Z(91) = \frac{91 - 55}{18} = 2.000 \] Now, we know that about 68% of the data falls within 1 standard deviation of the mean, and about 95% falls within 2 standard deviations. Therefore, we want to find the probability between these two Z-scores. The area between \( Z = 1 \) and \( Z = 2 \): - The area to the left of \( Z = 2 \) is about 0.9772 (from Z-tables). - The area to the left of \( Z = 1 \) is about 0.8413. Now, subtract the two probabilities: \[ P(73 < X < 91) = P(Z < 2) - P(Z < 1) \] \[ P(73 < X < 91) = 0.9772 - 0.8413 = 0.1359 \] Thus, the probability that \( X \) is between 73 and 91 is approximately **0.136**, rounded to the nearest thousandth.