The life spans of a species of fruit fly have a bell-shaped distribution, with a mean of 31 days and a standard deviation of 5 days. (a) The life spans of three randomly selected fruit flies are 34 days, 30 days, and 47 days. Find the \( z \)-score that corresponds to each life span. Determine whether any of these life spans are unusual. (b) The life spans of three randomly selected fruit flies are 36 days, 41 days, and 21 days. Using the Empirical Rule, find the percentile that corresponds to each life span. (a) The z-score corresponding a life span of 34 days is 60 . (Type an integer or a decimal rounded to two decimal places as needed.) The z-score corresponding a life span of 30 days is (Type an integer or a decimal rounded to two decimal places as needed.)

To solve the problem, we will follow these steps: ### Part (a): Finding the \( z \)-scores The formula for calculating the \( z \)-score is: \[ z = \frac{(X - \mu)}{\sigma} \] where: - \( X \) is the value (life span in this case), - \( \mu \) is the mean, - \( \sigma \) is the standard deviation. Given: - Mean (\( \mu \)) = 31 days - Standard deviation (\( \sigma \)) = 5 days We will calculate the \( z \)-scores for the life spans of 34 days, 30 days, and 47 days. 1. For \( X = 34 \): \[ z_{34} = \frac{(34 - 31)}{5} \] 2. For \( X = 30 \): \[ z_{30} = \frac{(30 - 31)}{5} \] 3. For \( X = 47 \): \[ z_{47} = \frac{(47 - 31)}{5} \] Let's calculate these \( z \)-scores. Calculate the value by following steps: - step0: Calculate: \(\frac{\left(34-31\right)}{5}\) - step1: Remove the parentheses: \(\frac{34-31}{5}\) - step2: Subtract the numbers: \(\frac{3}{5}\) Calculate or simplify the expression \( (47 - 31) / 5 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(47-31\right)}{5}\) - step1: Remove the parentheses: \(\frac{47-31}{5}\) - step2: Subtract the numbers: \(\frac{16}{5}\) Calculate or simplify the expression \( (30 - 31) / 5 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(30-31\right)}{5}\) - step1: Remove the parentheses: \(\frac{30-31}{5}\) - step2: Subtract the numbers: \(\frac{-1}{5}\) - step3: Rewrite the fraction: \(-\frac{1}{5}\) The calculated \( z \)-scores for the life spans are as follows: 1. For a life span of 34 days: \[ z_{34} = 0.6 \] 2. For a life span of 30 days: \[ z_{30} = -0.2 \] 3. For a life span of 47 days: \[ z_{47} = 3.2 \] ### Determining Unusual Life Spans A life span is considered unusual if its \( z \)-score is less than -2 or greater than 2. - \( z_{34} = 0.6 \) (not unusual) - \( z_{30} = -0.2 \) (not unusual) - \( z_{47} = 3.2 \) (unusual) ### Part (b): Finding Percentiles Using the Empirical Rule The Empirical Rule states that for a normal distribution: - Approximately 68% of the data falls within 1 standard deviation of the mean. - Approximately 95% falls within 2 standard deviations. - Approximately 99.7% falls within 3 standard deviations. Given: - Mean (\( \mu \)) = 31 days - Standard deviation (\( \sigma \)) = 5 days We will calculate the percentiles for the life spans of 36 days, 41 days, and 21 days. 1. For \( X = 36 \): - \( z_{36} = \frac{(36 - 31)}{5} = 1 \) (within 1 standard deviation) 2. For \( X = 41 \): - \( z_{41} = \frac{(41 - 31)}{5} = 2 \) (within 2 standard deviations) 3. For \( X = 21 \): - \( z_{21} = \frac{(21 - 31)}{5} = -2 \) (within 2 standard deviations) Now, we will determine the percentiles for these \( z \)-scores. - \( z = 1 \) corresponds to the 84th percentile (50% + 34%). - \( z = 2 \) corresponds to the 97.5th percentile (50% + 47.5%). - \( z = -2 \) corresponds to the 2.5th percentile (50% - 47.5%). Thus, the percentiles are: - For 36 days: 84th percentile - For 41 days: 97.5th percentile - For 21 days: 2.5th percentile ### Summary of Results - **Part (a)**: - \( z \)-score for 34 days: 0.6 - \( z \)-score for 30 days: -0.2 - \( z \)-score for 47 days: 3.2 (unusual) - **Part (b)**: - Percentile for 36 days: 84th percentile - Percentile for 41 days: 97.5th percentile - Percentile for 21 days: 2.5th percentile