Let $\mathrm{x}$ be a continuous random variable that is normally distributed with mean $\mu =24$ and standard deviation $\sigma =2$. Using the accompanying standard normal distribution table, find $P(26\le x\le 28)$.
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The probability is (Round to four decimal places as needed.)

Step 1 ：Let $x$ be a continuous random variable that is normally distributed with mean $\mu =24$ and standard deviation $\sigma =2$. We are asked to find the probability that $x$ falls between 26 and 28, i.e., $P(26\le x\le 28)$.

Step 2 ：To solve this, we first convert the values 26 and 28 to z-scores, which are standard deviations from the mean. The formula for a z-score is $(x-\mu )/\sigma$.

Step 3 ：Using this formula, we find that the z-score for 26 is 1.0 and the z-score for 28 is 2.0.

Step 4 ：We then use the standard normal distribution table to find the probabilities associated with these z-scores. The probability that $x$ is between 26 and 28 is the same as the probability that $z$ is between the z-score of 26 and the z-score of 28.

Step 5 ：From the standard normal distribution table, we find that the probability associated with a z-score of 1.0 is approximately 0.8413 and the probability associated with a z-score of 2.0 is approximately 0.9772.

Step 6 ：We subtract the smaller probability from the larger one to get the final answer. So, $P(26\le x\le 28)=P(z_2)-P(z_1)=0.9772-0.8413=0.1359$.

Step 7 ：Final Answer: The probability that a normally distributed random variable falls between 26 and 28 is approximately $\boxed{{\displaystyle 0.1359}}$.