Question

The life spans of a species of fruit fly have a​ bell-shaped distribution, with a mean of 36 days and a standard deviation of 5 days. ​(a) The life spans of three randomly selected fruit flies are 38 ​days, 32 ​days, and 48 days. Find the​ z-score that corresponds to each life span. Determine whether any of these life spans are unusual.

(a) The​ z-score corresponding a life span of 38 days =

   The​ z-score corresponding a life span of 32 days =

   The​ z-score corresponding a life span of 48 days =

b.) Select all of the life spans that are unusual:

   38 days

   32 days

   48 days

   None of the life spans are unusual.

c.) Determine the percentiles using the Empirical Rule:

The life spans of three randomly selected fruit flies are

46 days, 26 days, and 51 days. Using the Empirical​ Rule, find the percentile that corresponds to each life span.

The 46 day fruit fly corresponds to the___th percentile.

The 26 day fruit fly corresponds to the___th percentile.

The 51 day fruit fly corresponds to the___th percentile.

Answer 1

z-score for 38 days =

z-score for 32 days =

z-score for 48 days =

b) 48 days is the answer because when z > 2 or z < -2 then the event is unusual.

c) Empirical rule is: 68-95-99.7

i) 46 = 36 + 2*5 = 2 sd above from the mean.

So, 97.5% data are below the 46 days. So, this is 98 th percentile.

ii) 26 = 36 - 2*5 = 2 sd below from mean

So, 97.5% data are above the 26 days. So, this is 2 nd percentile.

iii) 51 = 36 + 3*5 = 3 sd above the mean.

So, 99.85% data are less than 51 days. So, this is 100 th percentile.

Please comment if any doubt. Thank you.