On average, indoor cats live to 15 years old with a standard deviation of 2.3 years. Suppose that the distribution is normal. Let X= the age at death of a randomly selected indoor cat. Round answers to 4 decimal places where possible.
c. The middle $30 % of indoor cats' age of death lies between what two numbers?
Answer
Final Answer: The middle 30% of indoor cats' age of death lies between \(\boxed{14.1138}\) years and \(\boxed{15.8862}\) years.
Step 1 ：Given that the average lifespan of indoor cats is 15 years with a standard deviation of 2.3 years, and the distribution is normal. Let X represent the age at death of a randomly selected indoor cat.
Step 2 ：We are asked to find the middle 30% of indoor cats' age of death. Since the distribution is symmetric, the middle 30% will be split evenly on either side of the mean. Therefore, we need to find the ages that correspond to the 35th percentile and the 65th percentile.
Step 3 ：First, we find the z-scores that correspond to the 35th percentile and the 65th percentile. The z-scores are approximately -0.3853 and 0.3853 respectively.
Step 4 ：Next, we convert these z-scores back into years using the given mean and standard deviation. The formula to convert z-scores to values is \(X = \mu + Z \sigma\), where \(\mu\) is the mean, \(Z\) is the z-score, and \(\sigma\) is the standard deviation.
Step 5 ：Substituting the given values into the formula, we get the ages corresponding to the 35th percentile and the 65th percentile as approximately 14.1138 years and 15.8862 years respectively.
Step 6 ：Final Answer: The middle 30% of indoor cats' age of death lies between \(\boxed{14.1138}\) years and \(\boxed{15.8862}\) years.
