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A normal random variable \(x\) has an unknown mean \(\mu\) and standard deviation \(\sigma=2\). If the probability that \(x\) exceeds 7.5 is \(.8023,\) find \(\mu\).

Short Answer

Expert verified
Answer: The mean of the distribution is 9.24.

Step by step solution

01

Identify the information given

We have the following data given in the question: - The standard deviation (σ) is 2. - The probability that the normal random variable (x) exceeds 7.5 is 0.8023. Our goal is to find the mean (μ) of the distribution.
02

Use the z-score formula

The z-score represents the number of standard deviations a value is away from the mean of the distribution. It can be calculated by using the following formula: \(Z = \frac{X - \mu}{\sigma}\) In this problem, we must find the z-score that corresponds to the area 0.8023 on the right side of the distribution (since it represents the probability that x exceeds 7.5). This means that we need to look for the area on the left side, which is given by \(1 - 0.8023 = 0.1977\).
03

Find the z-score

Using a standard normal distribution table, we can find the z-score that corresponds to an area to the left of 0.1977. The table value closest to this is 0.197664. So, the z-score is -0.87 (keep in mind that the table represents the area to the left, so a negative z-score means the value is below the mean).
04

Plug the z-score, x value, and σ into the formula

Now that we know the z-score (-0.87), we can use it along with the x value (7.5) and the standard deviation (2) to find the mean (μ) using the formula: \(-0.87 = \frac{7.5 - \mu}{2}\)
05

Solve for μ

By solving for μ (the mean) in the equation, we get: \(-0.87 \times 2 = 7.5 - \mu\) \(-1.74 = 7.5 - \mu\) \(\mu = 7.5 + 1.74\) \(\mu = 9.24\) The mean (μ) of the normal random variable is 9.24.

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