\( X \) is a normally distributed random variable with mean 68 and standard deviation 4 . What is the probability that \( X \) is between 56 and 64 ? Use the \( 0.68-0.95-0.997 \) rule and write your answer as a decimal. Round to the nearest thousandth if necessary.

To find the probability that \( X \) falls between 56 and 64, we can use the properties of the normal distribution along with the 68-95-99.7 rule. First, we calculate how many standard deviations away 56 and 64 are from the mean of 68. 1. For 56: \( z_1 = \frac{56 - 68}{4} = \frac{-12}{4} = -3 \) 2. For 64: \( z_2 = \frac{64 - 68}{4} = \frac{-4}{4} = -1 \) Now, referring to the 68-95-99.7 rule, about 68% of the data lies within 1 standard deviation from the mean, and about 99.7% lies within 3 standard deviations. - The range from \( z = -3 \) to \( z = -1 \) encompasses a portion of the area in the tail. The area to the left of \( z = -3 \) can be approximated as 0.001 (or 0.1%), indicating almost all the data is to the right. - The area between \( z = -1 \) and the mean (which corresponds to 68%) can be approximately calculated as half of that (34%) plus the area from the left of the mean to this point (~50%). Thus, the probability \( P(56 < X < 64) \) is given by: - Area from \( z = -3 \) to \( z = -1 \): Area left of -3 is approximately 0.001 Area left of -1 is around 0.1587, and we need to consider the area between these two values: \[ P(-3 < z < -1) = P(z < -1) - P(z < -3) \approx 0.1587 - 0.001 = 0.1577 \] So, rounding to the nearest thousandth, the probability that \( X \) is between 56 and 64 is approximately \( 0.158 \).