Question

the life spans of a species of fruit fly have a bell - shaped distribution, with a mean of 32 days and a standard deviation of 5 days.
(a) the life spans of three randomly selected fruit flies are 35 days, 29 days, and 44 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 37 days, 27 days, and 22 days. using the empirical rule, find the percentile that corresponds to each life span.
(type an integer or a decimal rounded to two decimal places as needed.)
the z - score corresponding a life span of 29 days is - 0.6.
(type an integer or a decimal rounded to two decimal places as needed.)
the z - score corresponding a life span of 44 days is 2.4.
(type an integer or a decimal rounded to two decimal places as needed.)
select all of the life spans that are unusual.
a. 44 days
b. 29 days
c. 35 days
d. none of the life spans are unusual.
(b) determine the percentiles using the empirical rule.
the 37 day fruit fly corresponds to a percentile of 84.
the 27 day fruit fly corresponds to a percentile of 16.
the 22 day fruit fly corresponds to a percentile of 3.
(round to the nearest whole number as needed.)

Explanation:

Step1: Calculate z - score formula

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. Given \(\mu = 32\) and \(\sigma=5\).
For \(x = 35\):
\(z=\frac{35 - 32}{5}=\frac{3}{5}=0.6\)
For \(x = 29\):
\(z=\frac{29 - 32}{5}=\frac{- 3}{5}=-0.6\)
For \(x = 44\):
\(z=\frac{44 - 32}{5}=\frac{12}{5}=2.4\)
A data point is considered unusual if \(|z|>2\). Since \(|2.4|>2\), the life - span of 44 days is unusual.

Step2: Use the Empirical Rule for percentiles

The Empirical Rule states that for a bell - shaped (normal) distribution:

• Approximately 68% of the data lies within \(\mu\pm\sigma=(32 - 5,32 + 5)=(27,37)\)
• Approximately 95% of the data lies within \(\mu\pm2\sigma=(32-10,32 + 10)=(22,42)\)
• Approximately 99.7% of the data lies within \(\mu\pm3\sigma=(32-15,32 + 15)=(17,47)\)

For \(x = 37\):
Since \(37=\mu+\sigma\), and about 68% of the data is within \(\mu\pm\sigma\), the percentage of data less than \(37\) is \(\frac{100\% - 68\%}{2}+68\%=84\%\) (because the normal distribution is symmetric). So the 37 - day life - span corresponds to the 84th percentile.

For \(x = 27\):
Since \(27=\mu-\sigma\), the percentage of data less than \(27\) is \(\frac{100\% - 68\%}{2}=16\%\). So the 27 - day life - span corresponds to the 16th percentile.

For \(x = 22\):
Since \(22=\mu - 2\sigma\), the percentage of data less than \(22\) is \(\frac{100\% - 95\%}{2}=2.5\%\approx3\%\) (rounded to the nearest whole number). So the 22 - day life - span corresponds to the 3rd percentile.

Answer:

(a) The z - scores for 35 days, 29 days, and 44 days are \(0.6\), \(-0.6\), and \(2.4\) respectively. The unusual life - span is 44 days.
(b) The percentiles for 37 days, 27 days, and 22 days are 84, 16, and 3 respectively.