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12:38 pm
suppose the random variable, x, is distributed normal with mean 5.28 and standard deviation 0.89. suppose the probability that x is greater than the value x is equal to 0.33, that is, p(x>x) = 0.33. what is x?
x=
round to 2 decimal places.
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Explanation:

Step1: Find the z - score

Given \(P(X > x)=0.33\), then \(P(X\leq x)=1 - 0.33=0.67\).
Using the standard normal distribution table (or a calculator with a normal - distribution inverse function), the z - score \(z\) corresponding to a cumulative probability of \(0.67\) is approximately \(z = 0.44\) (using a standard normal table or a calculator like the invNorm function in a TI - 84: invNorm\((0.67)\)).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 5.28\) (mean) and \(\sigma=0.89\) (standard deviation).
We know \(z = 0.44\), and we want to solve for \(x\).
Rearrange the formula to \(x=\mu+z\sigma\).
Substitute the values: \(x = 5.28+0.44\times0.89\).
First, calculate \(0.44\times0.89 = 0.3916\).
Then \(x=5.28 + 0.3916=5.6716\).

Answer:

\(5.67\)