Researchers collected information on the body parts of a new species of frog. The thumb length for the female frog has a mean of 7.73 mm and a standard deviation of 0.65 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 7.8 mm and 5.9 mm. Round your answers to two decimal places. a. Find the standardized version of x. zequals StartFraction x minus 7.73 Over 0.65 EndFraction (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 7.8 mm and 5.9 mm. Round your answers to two decimal places. Determine and interpret the z-score for a thumb length of 7.8 mm.
Solution:
We are given mean = 7.73, SD = 0.65
a. Find the standardized version of x.
Standardized version of x is given as below:
Z = (X – mean) / SD
Z = (X – 7.73) / 0.65
b. Determine and interpret the z-scores for thumb lengths of 7.8 mm and 5.9 mm.
First of all we have to find z-score for x = 7.8
Z = (X – mean) / SD
Z = (7.8 - 7.73) / 0.65
Z = 0.107692
Z = 0.11
Now, we have to find the z-score for x = 5.9
Z = (X – mean) / SD
Z = (5.9 - 7.73) / 0.65
Z = -2.81538
Z = -2.82
Determine and interpret the z-score for a thumb length of 7.8 mm.
The z-score for the thumb length of 7.8 mm is given as 0.11 and this z-score is greater than 0, which indicate that more than 50% of the thumb lengths are less than 7.8 mm.
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