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 37 ​days, 27 ​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 22 ​days, 27 ​days, and 42 days. Using the Empirical​ Rule, find the percentile that corresponds to each life span.

Respuesta :

(a) z score (37)  = 1

    z score (27) = -1

    z score (44)  = 2.4

(b) Ф(-2) = 1 - 0.97 = 3%

     Ф(-1) = 1 - 0.84 = 16%

     Ф(2) = 0.97 = 97%

Explanation:

Given -

Mean of life span = 32 days

Standard deviation of life span = 5 days

Life span of 3 random selected fruit flies are = 37, 27 and 44

(a) Z score = data point - mean / standard deviation

z score (37)  = 37 - 32 / 5 = 1

z score (27) = 27 - 32 / 5 = -1

z score (44) = 44 - 32 / 5 = 2.4

A positive z-score says the life span is above average.

A negative z-score says the life span is below average.

(b) Life span of 3 randomly selected fruit flies = 22, 27, 42

z score (1) = 22-32/5 = -2

z score (2) = 27-32/5 = -1

z score (3) = 42-32/5 = 2

From the z score table, the value of phi is taken

z(22) = -2

Ф(-2) = 1 - 0.97 = 3%

z(27) = -1

Ф(-1) = 1 - 0.84 = 16%

z(42) = 2

Ф(2) = 0.97 = 97%