Math Problem Statement

A random variable is known to be normally distributed with a standard deviation of 0.5. For a an x value of 6.1425, what should be the population mean so that the probability of a value from this distribution exceeding 6.1425 is at most 0.05. A. 3.98 B. 4.47 C. 6.02  D. 5.32

Solution

To find the population mean μ\mu such that the probability of a value exceeding 6.1425 is at most 0.05, we can use the properties of the normal distribution.

Step-by-step solution:

  1. Interpret the problem: The problem states that the random variable follows a normal distribution with a standard deviation σ=0.5\sigma = 0.5. The goal is to find the mean μ\mu such that the probability of the random variable XX exceeding x=6.1425x = 6.1425 is at most 0.05.

    Mathematically, we are looking for: P(X>6.1425)=0.05P(X > 6.1425) = 0.05 which means we are interested in the upper tail of the normal distribution.

  2. Use the Z-score: To find this probability, we first need to convert the value of 6.1425 to a Z-score. The Z-score formula is: Z=xμσZ = \frac{x - \mu}{\sigma} where x=6.1425x = 6.1425, μ\mu is the mean, and σ=0.5\sigma = 0.5.

  3. Find the Z-score corresponding to the 95th percentile: Since we want the probability to be at most 0.05, we need to find the Z-score that corresponds to the cumulative probability of 0.95 (because 0.05 is in the upper tail, and the cumulative probability to the left of the Z-value should be 0.95).

    From standard Z-tables or using statistical software, the Z-score corresponding to a cumulative probability of 0.95 is approximately Z=1.645Z = 1.645.

  4. Set up the equation: Using the Z-score formula, we have: 1.645=6.1425μ0.51.645 = \frac{6.1425 - \mu}{0.5}

  5. Solve for μ\mu: Multiply both sides by 0.5: 1.645×0.5=6.1425μ1.645 \times 0.5 = 6.1425 - \mu 0.8225=6.1425μ0.8225 = 6.1425 - \mu Rearranging: μ=6.14250.8225\mu = 6.1425 - 0.8225 μ=5.32\mu = 5.32

Thus, the population mean μ\mu should be 5.32.

Final Answer:

D. 5.32


Would you like a detailed explanation of any of these steps?

Here are 5 related questions to explore:

  1. How do we calculate the Z-score for other cumulative probabilities?
  2. What is the interpretation of a Z-score in a normal distribution?
  3. How can we apply the same method to find the mean for a different probability, say 0.01 or 0.10?
  4. How do we solve for the Z-score when given a specific probability threshold?
  5. How does the standard deviation affect the Z-score and the corresponding value?

Tip: Always remember that the Z-score tells you how many standard deviations a given value is away from the mean!

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Z-score
Probability
Statistical Inference

Formulas

Z = (x - μ) / σ
P(X > x) = 0.05

Theorems

Z-table
Properties of the Normal Distribution

Suitable Grade Level

Grades 10-12