\( X \) is a normally distributed random variable with mean 38 and standard deviation 13 . What is the probability that \( X \) is between 51 and 77 ? Use the \( 0.68-0.95-0.997 \) rule and write your answer as a decimal. Round to the nearest thousandth if necessary.
Solución de tutoría real
Responder
Solución
Revisado y aprobado por el equipo de tutoría de UpStudy
The Deep Dive
To find the probability that \( X \) is between 51 and 77, first, we need to find how many standard deviations away these values are from the mean. The mean \( \mu \) is 38 and the standard deviation \( \sigma \) is 13. For \( X = 51 \): \[ Z_1 = \frac{51 - 38}{13} = \frac{13}{13} = 1 \] For \( X = 77 \): \[ Z_2 = \frac{77 - 38}{13} = \frac{39}{13} = 3 \] Now, according to the \( 0.68-0.95-0.997 \) rule: - About 68% of the data lies within 1 standard deviation (from \( 38 - 13 \) to \( 38 + 13 \)). - About 95% lies within 2 standard deviations (from \( 38 - 2 \times 13 \) to \( 38 + 2 \times 13 \)), which translates to the interval from \( 12 \) to \( 64 \). - About 99.7% lies within 3 standard deviations (from \( 38 - 3 \times 13 \) to \( 38 + 3 \times 13 \)), which translates to the interval from \( -1 \) to \( 77 \). Now, we are interested in the probability between \( 51 \) (which is 1 standard deviation away) and \( 77 \) (which is 3 standard deviations away). Since we know that: - From the mean (38) to 1 standard deviation above (51) is 34% (half of 68%). - From the mean to 3 standard deviations above (77) is 49.85% (half of 99.7%). Therefore, the probability that \( X \) is between 51 and 77 is: \[ 0.34 + 0.497 = 0.84 \] Thus, the final probability that \( X \) is between 51 and 77 is approximately \( 0.840 \).