Ask Question
28 April, 09:15

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.

+1
Answers (1)
  1. 28 April, 09:25
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Fawns Fawns between 1 and 5 months old in Mesa Verde National Park have a body weight that is approximately normally distributed with mean ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers