QUESTION 10 Assume that $\mathrm{X}$ has a normal distribution, and find the indicated probability. The mean is $\mu=60.0$ and the standard deviation is $\sigma=4.0$. Find the probability that $X$ is less than 53.0. 0.9599 0.5589 0.0802 0.0401

Step 1 ：Given that X has a normal distribution with a mean (μ) of 60.0 and a standard deviation (σ) of 4.0, we are asked to find the probability that X is less than 53.0.

Step 2 ：First, we calculate the Z-score for X=53.0 using the given mean and standard deviation. The Z-score is calculated as \(Z = \frac{X - \mu}{\sigma}\).

Step 3 ：Substituting the given values, we get \(Z = \frac{53.0 - 60.0}{4.0} = -1.75\).

Step 4 ：Next, we use a standard normal distribution table to find the probability that X is less than 53.0. The probability corresponding to Z=-1.75 is approximately 0.0401.

Step 5 ：This means that there is a 4.01% chance that a randomly selected value from this normal distribution will be less than 53.0.

Step 6 ：Final Answer: The probability that X is less than 53.0 is \(\boxed{0.0401}\).