Question

The life spans of a species of fruit fly have a​ bell-shaped distribution, with a mean of 34 days and a standard deviation of 4 days. ​(a) The life spans of three randomly selected fruit flies are 36 ​days, 31 ​days, and 46 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 26 ​days, 30 ​days, and 46 days. Using the Empirical​ Rule, find the percentile that corresponds to each life span.

Answer 1

Mean = 34 days

Standard deviation = 4 days

a) Z = (X - mean)/standard deviation

Z score corresponding to 36 = (36 - 34)/4 = 0.5

Z score corresponding to 31 = (31 - 34)/4 = -0.75

Z score corresponding to 46 = (46 - 34)/4 = 3

A value is unusual if the z score is below -2 or above 2

So, the unusual life span is 46 days

b) According to the empirical rule, 68%, 95% and 99.7% of data values lie within 1, 2 and 3 standard deviations of mean.

26 is 2 standard deviations below mean. 95% is between 26 and 42

95/2 = 47.5% is between 26 and 34

50% is above 34

Percentile rank of 26 = 100 - (47.5 + 50) = 2.5 percentile

30 is 1 standard deviation from mean

Percentage of data between 30 and 34 = 68/2 = 34%

Percentile rank of 30 = 100 - (34 + 50) = 16 percentile

46 is 3 standard deviations above mean.

Percentage of data between 34 and 46 = 99.7/2 = 49.85%

Percentage of data below 34 = 50%

Perentile rank of 34 = 50 + 49.85 = 99.85 percentile.