Math Problem Statement

Company XYZ know that replacement times for the quartz time pieces it produces are normally distributed with a mean of 11.6 years and a standard deviation of 1.3 years. Find the probability that a randomly selected quartz time piece will have a replacement time less than 8 years? P(X < 8 years) = ? If the company wants to provide a warranty so that only 2.9% of the quartz time pieces will be replaced before the warranty expires, what is the time length of the warranty?

Solution

Let's break down the question:

  1. **Given dataLet's break down the question:

  2. **Given **

    • The replacement times are normally distributed.
    • Mean μ=11.6\mu = 11.6 years.
    • Standard deviation σ=1.3\sigma = 1.3 years.

    First Part: Find the probability that a randomly selected quartz time piece has a replacement time less than 8 years.

    This is asking for P(X<8)P(X < 8), where XX is the normal random variable representing the replacement time.

    To solve this:

    • We need to calculate the Z-score using the formula:

    Z=XμσZ = \frac{X - \mu}{\sigma}

    For X=8X = 8:

    Z=811.61.3=3.61.32.769Z = \frac{8 - 11.6}{1.3} = \frac{-3.6}{1.3} \approx -2.769

    Now, we can look up the Z-score in the standard normal distribution table, or use a calculator. The Z-score of 2.769-2.769 corresponds to a cumulative probability of approximately 0.0028.

    So, the probability that a replacement time is less than 8 years is about 0.0028.

    Second Part: Warranty Calculation

    The company wants to set the warranty such that only 2.9% (or 0.029) of quartz timepieces will be replaced before the warranty expires. This is the area to the left of the desired warranty time in the normal distribution.

    • First, find the Z-score that corresponds to the 2.9% left-tail probability. Using a Z-table or calculator, the Z-score for 0.029 is approximately 1.88-1.88.

    • Now use the Z-score formula in reverse to find the corresponding warranty time, XX:

    Z=XμσZ = \frac{X - \mu}{\sigma}

    Solving for XX:

    X=Zσ+μX = Z \cdot \sigma + \mu

    Plugging in the values:

    X=(1.88)1.3+11.62.444+11.6=9.156X = (-1.88) \cdot 1.3 + 11.6 \approx -2.444 + 11.6 = 9.156

    So, the warranty should be approximately 9.2 years (rounding to 1 decimal place).


Let me know if you'd like further explanation or have more questions!

Here are five related questions you might find useful:

  1. How do we interpret a Z-score in the context of a normal distribution?
  2. What are some common Z-scores and their corresponding probabilities?
  3. How does changing the standard deviation affect the probability in a normal distribution?
  4. How can you determine the warranty period for different replacement percentages?
  5. What are some methods to approximate probabilities when Z-tables are not available?

Tip: Always remember to check whether the normal distribution is appropriate for the data by considering skewness or kurtosis before applying Z-scores.

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Math Problem Analysis

Mathematical Concepts

Probability
Normal Distribution
Z-scores

Formulas

Z = (X - μ) / σ
P(X < x) = Z-score probability
X = Z * σ + μ

Theorems

Standard Normal Distribution

Suitable Grade Level

Grades 10-12