Researchers collected information on the body parts of a new species of frog. Th

Question

Researchers collected information on the body parts of a new species of frog. The thumb length for the female frog has a mean of 6.46 mm and a standard deviation of 0.56 mm. Let x denote thumb length for a female specimen. a. Find the standardized version of x. b. Determine and interpret the z-scores for thumb lengths of 7.4 mm and 5.3 mm. Round your answers to two decimal places. a. Find the standardized version of x (Do not simplify. Use integers or decimals for any numbers in the expression. Do not round.) b. Determine and interpret the z-scores for thumb lengths of 74 mm and 5.3 mm. Round your answers to two decimal places. Determine and interpret the z-score for a thumb length of 7.4 mm. The z-score for a thumb length of 7.4 is This is standard deviations the mea n thumb lengh. (Round to two decimal places as needed.) Determine and interpret the z-score for a thumb length of 5.3 mm. The z-score for a thumb length of 5.3 IS This is standard deviations the mean thumb length. (Round to two decimal places as needed.)

Explanation / Answer

The params of the normal dsitribution are given below:

Mean = 6.46

Stdev = .56

We have to use them to solve the problem

a.
Z = x-Mean / Stdev = (x-6.46)/.56

b.
Z score for 7.4 is (7.4-6.46)/.56 = 1.68
This is 1.68 standard deviations above the mean thumb length

Z score for 5.3 is (5.3-6.46)/.56 = -2.07
This is 2.07 standard deviations below the mean thumb length

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