Suppose $x$ is a normally distributed random variable with $\mu=33$ and $\sigma=4$. Find a value $x_{0}$ of the random variable $x$. a. $P\left(x \geq x_{0}\right)=5$ b. $P\left(xx_{0}\right)=10$ d. $P\left(x>x_{0}\right)=.95$ Click here to view a table of areas under the standardized normal curve. a. $x_{0}=\square$ (Round to the nearest hundredth as needed.)

Step 1 ：The problem is asking to find the value of \(x_{0}\) for different probabilities under a normal distribution with mean \(\mu=33\) and standard deviation \(\sigma=4\).

Step 2 ：To solve this problem, we need to use the concept of Z-score in statistics. The Z-score is a measure of how many standard deviations an element is from the mean. We can calculate the Z-score using the formula: \(Z = \frac{X - \mu}{\sigma}\)

Step 3 ：We can find the Z-score corresponding to the given probabilities from the Z-table. Then, we can find the value of \(x_{0}\) using the formula: \(X = Z * \sigma + \mu\)

Step 4 ：For a, the Z-score is -1.6448536269514722. So, \(x_{0} = -1.6448536269514722 * 4 + 33 = 39.579414507805886\)

Step 5 ：For b, the Z-score is -1.6448536269514722. So, \(x_{0} = -1.6448536269514722 * 4 + 33 = 25.16014406183978\)

Step 6 ：For c, the Z-score is -1.6448536269514722. So, \(x_{0} = -1.6448536269514722 * 4 + 33 = 38.1262062621784\)

Step 7 ：For d, the Z-score is -1.6448536269514722. So, \(x_{0} = -1.6448536269514722 * 4 + 33 = 26.42058549219411\)

Step 8 ：Rounding to the nearest hundredth, we get the final answers: \(\boxed{x_{0}=39.58}\) for a, \(\boxed{x_{0}=25.16}\) for b, \(\boxed{x_{0}=38.13}\) for c, and \(\boxed{x_{0}=26.42}\) for d.