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.

(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%

solvent

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