Step 1 :Given that the average lifespan of indoor cats is 15 years with a standard deviation of 2.3 years, we are asked to find the probability that an indoor cat dies when it is between 10.7 and 13.1 years old. We assume that the distribution is normal.
Step 2 :First, we need to calculate the z-scores for the ages 10.7 and 13.1 years. The z-score is a measure of how many standard deviations an element is from the mean. The formula for calculating the z-score is \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the value from the dataset, \(\mu\) is the mean and \(\sigma\) is the standard deviation.
Step 3 :Substituting the given values into the formula, we get \(Z_1 = \frac{10.7 - 15}{2.3} = -1.8696\) and \(Z_2 = \frac{13.1 - 15}{2.3} = -0.8261\).
Step 4 :Next, we use the standard normal distribution table or a function that gives the cumulative distribution function to find the probabilities corresponding to these z-scores. The probabilities we get are 0.0308 and 0.2044 respectively.
Step 5 :The probability that an indoor cat dies when it is between 10.7 and 13.1 years old is the difference between these two probabilities. So, the probability is \(0.2044 - 0.0308 = 0.1736\).
Step 6 :Final Answer: The probability that an indoor cat dies when it is between 10.7 and 13.1 years old is approximately \(\boxed{0.1736}\).