Fawns Fawns between 1 and 5 months old in Mesa Verde National Park have a body weight that is approximately normally distributed with mean m 5 27.2 kilograms and standard deviation s 5 4.3 kilograms (based on information from The Mule Deer of Mesa Verde National Park, by G. W. Mierau and J. L. Schmidt, Mesa Verde Museum Association). Let x be the weight of a fawn in kilograms. Convert each of the following x intervals to z intervals. (a) x 6 30 (b) 19 6 x (c) 32 6 x 6 35 Brase, Charles Henry. Understandable Statistics (p. 297). Cengage Learning. Kindle Edition.

a) x < 30 = z < 0.65

b) 19 < x = -1.91 < z

Or x > 19 = z > -1.91

c) 32 < x < 35 = 1.12 < z < 1.81

Step-by-step explanation:

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - xbar)/σ

x = the value to be standardized, that is, converted to z-score

xbar = mean = 27.2 kg

σ = standard deviation = 4.3 kg

a) For x < 30

We standardize the 30 kg weight first

z = (x - xbar)/σ = (30 - 27.2)/4.3 = 0.65

x < 30 = z < 0.65

b) For 19 < x

We standardize the 19 kg weight first

z = (x - xbar)/σ = (19 - 27.2)/4.3 = - 1.91

19 < x = - 1.91 < z

Or x > 19 = z > - 1.91

c) For 32 < x < 35

We standardize the 32 kg and 35 kg weights first

z = (x - xbar)/σ

z₁ = (32 - 27.2)/4.3 = 1.12

z₂ = (35 - 27.2)/4.3 = 1.81

32 < x < 35 = 1.12 < z < 1.81

is there supposed to be a diagram or a graph for this one? ?

step-by-step explanation: