The life spans of a species of fruit fly have a bell-shaped distribution, with a

Question

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 4 days. (a) The life spans of three randomly selected fruit flies are 34 days, 29 days and 43 days. Find the z-score that corresponds to each life span. Determine whether (b) The life spans of three randomly selected fruit flies are 43 days, 23 days, and 35 days using the Empirical Rule, find the percentile that corresponds to each. (a) The z-score corresponding a life span of 34 days is 0.75 (Type an integer or a decimal rounded to two decimal places as needed.) The z-score corresponding a life span of 29 days is -0.5. (Type an integer or a decimal rounded to two decimal places as needed.) The z-score corresponding a life span of 43 days is 3. (Type an integer or a decimal rounded to two decimal places as needed.) Select all of the life spans that are unusual. A. 34 days B. 43 days C. 29 days D None of the life spans unusual (b) Determine the percentiles using the Empirical Rule. The 43 day fruit fly corresponds to the the th percentile, The 23 day fruit fly corresponds to the th percentile, The 35 day fruit fly corresponds to the th percentile. (Type an integer or a decimal.)

Explanation / Answer

b)

For 43 days, 43 - 31 = 12 , 12/4 = 3 This means 43 lies at 3 sigma level from the mean .

Hence, Required percentile = 0.5 + 0.997/2 = 99.85 percentile

For 23 days, 23 - 31 = -8 , -8/4 = -2 This means 23 lies at 2 sigma to the left level from the mean .

Hence, Required percentile = 0.025 = 2.5percentile

For 35 days, 35 - 31 = 4 , 4/4 = 1 This means 35 lies at 1 sigma level to the right from the mean .

Hence, Required percentile = 0.5+ 0.68/2 = 84 percentile

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.