Questions: The probability that a new car of a certain model will require repairs during the warranty period is 0.15. A particular dealership sells 80 such cars. Let X be the number that will require repairs during the warranty period. Find the mean and standard deviation of X.

The probability that a new car of a certain model will require repairs during the warranty period is 0.15. A particular dealership sells 80 such cars. Let X be the number that will require repairs during the warranty period. Find the mean and standard deviation of X.

Transcript text: The probability that a new car of a certain model will require repairs during the warranty period is 0.15 . A particular dealership sells 80 such cars. Let $X$ be the number that will require repairs during the warranty period. Find the mean and standard deviation of $X$.

Solution

Solution Steps

Step 1: Define the Problem

We are given a scenario where the probability that a new car of a certain model will require repairs during the warranty period is $ p = 0.15 $. A dealership sells $ n = 80 $ such cars. We need to find the mean and standard deviation of the random variable $ X $, which represents the number of cars that will require repairs during the warranty period.

Step 2: Calculate the Mean

The mean $ \mu $ of a binomial distribution can be calculated using the formula: \[ \mu = n \cdot p \] Substituting the values: \[ \mu = 80 \cdot 0.15 = 12.0 \]

Step 3: Calculate the Variance

The variance $ \sigma^2 $ of a binomial distribution is given by: \[ \sigma^2 = n \cdot p \cdot q \] where $ q = 1 - p = 0.85 $. Thus, \[ \sigma^2 = 80 \cdot 0.15 \cdot 0.85 = 10.2 \]

Step 4: Calculate the Standard Deviation

The standard deviation $ \sigma $ is the square root of the variance: \[ \sigma = \sqrt{n \cdot p \cdot q} = \sqrt{10.2} \approx 3.19 \]