Question

The wingspans of cardinals follow a Normal distribution with a mean of 11 inches and standard deviation of 0.6 inches. a. Two cardinals, a female and male, were sitting in the bush outside my front window. Their wingspans were 10.8 and 12.3 respectively. Find and compare the z-scores for these two cardinals. Which cardinal has the more extreme wingspan? c. If we were to repeatedly catch and release samples of size 16 cardinals, what are the z-scores for two samples which had average wingspans of 10.8 for sample 1 and 12.3 for sample 2? How do these z-scores compare to the previous ones? d. Now, if we were to repeatedly catch and release samples of size 36 cardinals, what is the probability that a randomly selected sample average would fall between 10.8 and 12.3? Use StatCrunch to answer and include the output or graph created.

Answer 1

µ = 11, σ = 0.6

a) z-scores for 10.8 and 12.3

z = (X-µ)/σ = (10.8-11)/0.6 = -0.33

z = (X-µ)/σ = (12.3-11)/0.6 = 2.17

Male has the more extreme wingspan.

c)

µ = 11, σ = 0.6, n = 16

For sample 1:

z = (X̅-μ)/(σ/√n) = (10.8-11)/(0.6/√16) = -1.33

For sample 2:

z = (X̅-μ)/(σ/√n) = (12.3-11)/(0.6/√16) = 8.67

d)

µ = 11, σ = 0.6, n = 36

P(10.8 < X̅ < 12.3) =

= P( (10.8-11)/(0.6/√36) < (X-µ)/(σ/√n) < (12.3-11)/(0.6/√36) )

= P(-2 < z < 13)

= P(z < 13) - P(z < -2)

Using excel function:

= NORM.S.DIST(13, 1) - NORM.S.DIST(-2, 1)

= 0.9772