Step 1 :Assume the random variable $X$ is normally distributed with mean $\mu=50$ and standard deviation $\sigma=7$. We are asked to compute the probability $P(56 \leq X \leq 70)$.
Step 2 :First, we need to convert these values to z-scores, which give us the number of standard deviations away from the mean a value is. The formula for a z-score is $z = \frac{x - \mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation.
Step 3 :Using this formula, we find that the z-scores corresponding to 56 and 70 are $z_1 = \frac{56 - 50}{7} = 0.8571428571428571$ and $z_2 = \frac{70 - 50}{7} = 2.857142857142857$ respectively.
Step 4 :The probability that $X$ falls between 56 and 70 is then the probability that a standard normal random variable falls between these z-scores. This is given by $P(56 \leq X \leq 70) = P(z_1 \leq Z \leq z_2)$, where $Z$ is a standard normal random variable.
Step 5 :We can find these probabilities using a standard normal table or a statistical software. The probability is approximately $p = 0.19354560217368977$.
Step 6 :Final Answer: The probability that the random variable $X$ falls between 56 and 70 is approximately \(\boxed{0.1935}\).